The width of the rectangle is 3/2 cm and the length is 17 cm.
The formula for the area of a rectangle is A = L x W, where A is the area, L is the length, and W is the width. To solve this problem, we can rearrange the formula to solve for W: W = A / L. In this case, A = 6 cm and L = 2W + 11 cm. Substituting these values, we get W = 6 / (2W + 11). We can solve this equation by dividing both sides by 2W to get 1/W = 6 / (2W + 11) * 1/2W. Simplifying, we get 1/W = 3 / (11 + 2W). Subtracting 3/11 from both sides, we get -2/11 W = -3/11. Finally, we can solve for W by dividing both sides by -2/11 to get W = 3/2 cm.
Therefore, the width of the rectangle is 3/2 cm, and the length of the rectangle is 2W + 11 = 11 + 6 = 17 cm.
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Please help with #7 and #10 a-c
7. The speed of the car is approximately 73.3 miles per hour. 10. a. 7.07 ft/sec b. 9.87 feet from the center. c. the person on the outside is approximately 30 ft.
Describe Speed?It is a fundamental concept in physics and is used to describe the motion of objects in both everyday life and scientific contexts. In addition to speed, other related concepts include velocity (which includes direction) and acceleration (the rate at which an object's speed or velocity changes).
7. The circumference of the wheel is given by C = πd, where d is the diameter of the wheel. Therefore, the circumference of each wheel is:
C = πd = π(22 in) = 22π in
Since the wheels are rotating at 600 rpm (revolutions per minute), the distance covered by each wheel in one minute is:
distance = circumference × revolutions = 22π in/rev × 600 rev/min = 13,200π in/min
To convert this to miles per hour, we need to convert inches to miles and minutes to hours. There are 12 inches in 1 foot and 5280 feet in 1 mile, so there are 63360 inches in 1 mile. There are 60 minutes in 1 hour, so:
distance = 13,200π in/min = (13,200π in/min) × (1 ft/12 in) × (1 mi/5280 ft) × (60 min/1 hr) ≈ 73.3 mi/hr
Therefore, the speed of the car is approximately 73.3 miles per hour.
10. a) The linear speed of a point on the carousel is given by v = rω, where r is the distance from the center of the carousel and ω is the angular velocity in radians per second. Since the carousel makes 3 revolutions per minute, the angular velocity is:
ω = (3 rev/min) × (2π rad/rev) × (1 min/60 sec) = π/10 rad/sec
The linear speed of a person riding a horse that is 22.5 ft from the center is:
v = rω = (22.5 ft) × (π/10 rad/sec) ≈ 7.07 ft/sec
b) Let d be the distance of the person on the inside from the center. Since the carousel makes 3 revolutions per minute, the angular velocity is still ω = π/10 rad/sec. The linear speed of the person on the inside is given as 3.1 ft/sec, so:
v = rω
3.1 ft/sec = d × (π/10 rad/sec)
d = (3.1 ft/sec) × (10/π rad/sec) ≈ 9.87 ft
Therefore, the person on the inside is approximately 9.87 feet from the center.
c) The linear speed of a point on the carousel is proportional to its distance from the center. Therefore, the ratio of the linear speeds of the person on the outside and the inside is equal to the ratio of their distances from the center:
v_outside/v_inside = r_outside/r_inside
We know that the linear speed of the person on the inside is 3.1 ft/sec and the distance of the person on the inside is approximately 9.87 feet. Therefore:
v_outside/3.1 ft/sec = r_outside/9.87 ft
v_outside = (r_outside/9.87 ft) × (3.1 ft/sec)
We can solve for r_outside by using the fact that the carousel makes 3 revolutions per minute:
v_outside = r_outsideω
v_outside = r_outside × (3 rev/min) × (2π rad/rev) × (1 min/60 sec)
v_outside = r_outsideπ/10 rad/sec
Combining these equations, we get:
r_outsideπ/10 rad/sec = (r_outside/9.87 ft) × (3.1 ft/sec)
r_outside ≈ 30.9 ft
Therefore, the person on the outside is approximately 30.
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The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. With 95% confidence, what is the margin of error for the estimation of the population mean?
We estimated the margin of error for the population mean with 95% confidence using the given data and formula. The margin of error is approximately 3.32, providing a reasonable range for the true population mean.
To find the margin of error for estimating the population mean with 95% confidence, we can use the formula:
Margin of error = t(α/2) × (s/√n)
where t(α/2) is the critical value of the t-distribution with (n - 1) degrees of freedom and a significance level of α/2 (0.025 for 95% confidence), s is the sample standard deviation, and n is the sample size.
From the given data, we can calculate the sample mean and standard deviation as:
x = (10 + 8 + 12 + 15 + 13 + 11 + 6 + 5) / 8 = 9.75
s = √[((10-9.75)² + (8-9.75)² + (12-9.75)² + (15-9.75)² + (13-9.75)² + (11-9.75)² + (6-9.75)² + (5-9.75)²) / 7] = 3.36
Using a t-table or calculator, we can find the critical value t(0.025) with 7 degrees of freedom to be approximately 2.365.
Substituting these values into the formula, we get:
Margin of error = 2.365 × (3.36 / √(8)) ≈ 3.32
Therefore, with 95% confidence, the margin of error for estimating the population mean is approximately 3.32.
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5 shirts, 2 neckties, 4 pants, and 2 sweaters? Looks like the previous problem, but the dress-code has changed, and I am permitted to omit the tie, and if the weather is warm, omit the sweater
Thus, if the weather is warm, you can make a total of 20 combinations (5 shirts x 4 pants), and if the weather is cold, you can make a total of 40 combinations (5 shirts x 4 pants x 2 sweaters).
The total combination of clothing items you can wear is 5 shirts, 4 pants, and either 2 neckties or 2 sweaters. If the weather is warm, you can opt to omit the sweater and keep the two neckties. If the weather is cold, then you can omit the neckties and wear the two sweaters. To calculate the total number of clothing combinations you can make, you need to multiply the items you are allowed to wear. For example, if you are wearing a shirt and a pair of pants, you can make 5 x 4 = 20 combinations. If you are wearing a shirt, pants, and either a necktie or a sweater, you can make 5 x 4 x 2 = 40 combinations.
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Find T4,T7 and T9 for the geometric progression. 3,6,12,24
The given geometric progression's fourth, seventh, and ninth term is 24, 192, and 768 respectively.
In mathematics, a sequence known as a geometric progression (GP) is one in which each following term is generated by multiplying each preceding term by a constant integer, known as a common ratio. This progression is sometimes referred to as a pattern-following geometric sequence of numbers. Learn development in mathematics here as well. Here, each phrase is multiplied by the common ratio to generate the subsequent term, which is a non-zero value.
The given geometric progression: 3,6,12,24.......
The first term of GP is a = 3 and the Common ratio r = 2,
We know that the nth term of GP is,
[tex]T_n=ar^{n-1}[/tex]
We have to Find the 4, 7, and 9th terms of above mentioned GP,
[tex]T_4\\\\T_4=ar^{4-1}\\\\T_4=3*2^3\\\\T_4=24\\\\\\T_7\\\\T_7=3*2^{7-1}\\\\T_7=192\\\\\\T_9\\\\T_9=3*2^{9-1}\\\\T_9=768[/tex]
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Please ASAP Help
Will mark brainlest due at 12:00
Answer:
The answer is C.
Step-by-step explanation:
The midway point is (19/2, -6)
19/2 = 9.5
Hope this helps!
ps. you can type this in a calculator and ask it to find the midpoint, if that helps :)
Lilly Mae measures the length of the radius of her bedroom clock. She glued ribbon around the edge of the clock. What is the length of the ribbon lilly Mae glues around her clock? Use ~ 3.14
The length of the ribbon used by Lilly Mae is 15.7 inches.
The perimeter of the circle:A circle is a two-dimensional shape made up of points that are equidistant from a center point.
The perimeter of a circle is also known as its circumference and the formula for perimeter of circle is given by
C = 2πr
Where 'r' is the radius of the circle.
Here we have
Lilly Mae measures the length of the radius of her bedroom clock
From the figure, the Radius of the clock is 2.5 inch
Here the length of the ribbon around clock will equal the perimeter of the circular clock
From the formula, the Perimeter of the circle = 2πr
Length of the ribbon used = 2(3.14)(2.5) = 15.7 inches
Therefore,
The length of the ribbon used by Lilly Mae is 15.7 inches.
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On Cupcake Wars, the bakers must fill 1000 cupcakes in the final challenge!
CUPCAKE
WARS
If the cupcakes are cylinder shaped and have a radius of 1.5 in and a height of 3.01 in,
how many cubic inches are need to fill 1000 cupcakes?
Use pi=3.14
Round your answer to the nearest hundredth as needed.
V=
Answer:
21,265.65 in³
Step-by-step explanation:
Formula: Volume (V) = πr²h
V = πr²h
Slotting in values,
V = 3.14 x 1.5² x 3.01
V = 3.14 x 2.25 x 3.01
V = 21.26565 in³
But that is for 1 cupcake
To fill a thousand of them, you will need to multiply your answer by the number of cupcakes in total.
Therefore,
Answer = V x 1000
Answer = 21.26565 x 1000
Answer = 21,265.65 in³
The answer is already rounded to the nearest hundredth,
So they would need 21,265.65 cubic inches (in³) to fill 1000 cupcakes.
Please mark brainliest.
Thanks.
express each of the following as a single fraction 5y/2-2y/3
Answer:
Use a calculator for you to get an answer
From her eye, which stands 1.69 meters above the ground, Sadie measures the angle of elevation to the top of a prominent skyscraper to be 36 ∘ ∘ . If she is standing at a horizontal distance of 275 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.
We can use the tangent function to solve this problem. Let h be the height of the skyscraper.
First, we need to find the length of the adjacent side of the right triangle formed by Sadie, the base of the skyscraper, and the point where she is standing. This length is the horizontal distance between Sadie and the base of the skyscraper, which is 275 meters.
Next, we can use the tangent of the angle of elevation to find the ratio of the opposite side (the height of the skyscraper) to the adjacent side:
tan(36°) = h/275
Solving for h, we get:
h = 275 tan(36°)
Using a calculator, we find:
h ≈ 198.64 meters
Therefore, the height of the skyscraper is approximately 198.64 meters.
Answer: 201.49
Step-by-step explanation:
A school sports team contains 68 students. 33 do field events, 40 do track events, 23 do swimming, 14 do both field and track events. If 15 students do field events only and 10 do both swimming and track events, how many students do a) swimming only b)track events only c) all three events?
Let's label the regions in the Venn diagram as follows:
F: students who do only field events
T: students who do only track events
S: students who do only swimming events
FT: students who do both field and track events
ST: students who do both swimming and track events
FST: students who do all three events
From the information given in the problem, we can fill in some of the values in the Venn diagram:
F + FT + 15 = 33 (33 students do field events, and 15 do field events only)
T + FT + 14 = 40 (40 students do track events, and 14 do both field and track events)
S + ST + 10 = 23 (23 students do swimming, and 10 do both swimming and track events)
F + T + S + 2FT + ST + FST = 68 (there are 68 students in total)
We can simplify these equations to:
F + FT = 18
T + FT = 26
S + ST = 13
F + T + S + 2FT + ST + FST = 68
To solve for the remaining unknowns, we need to use some algebra. We can start by solving for FT:
F + FT = 18
T + FT = 26
Adding the two equations, we get:
F + T + 2FT = 44
Rearranging, we get:
FT = (44 - F - T) / 2
Now we can substitute this expression for FT into the equation for the total number of students:
F + T + S + 2FT + ST + FST = 68
F + T + S + 2((44 - F - T) / 2) + ST + FST = 68
F + T + S + 44 - F - T + ST + FST = 68
Simplifying, we get:
S + ST + FST = 26
Now we have two equations involving S, ST, and FST:
S + ST = 13
S + ST + FST = 26
We can solve for S and ST by subtracting the first equation from the second:
S + ST + FST = 26
(S + ST) + FST = 26
Substituting S + ST = 13:
13 + FST = 26
FST = 13
So there are 13 students who do all three events. Now we can use the equation S + ST = 13 to solve for S:
S + ST = 13
ST = 10 (given in the problem)
S = 13 - 10 = 3
So there are 3 students who do swimming only. Similarly, we can use the equation T + FT = 26 to solve for T:
T + FT = 26
FT = (44 - F - T) / 2
Substituting F + FT = 18:
T + (18 - F) / 2 = 26
Multiplying both sides by 2:
2T + 18 - F = 52
At December 31, bonds payable of $119,588,000 are outstanding the bonds pay 12% interest every September 30th in mature and installment of 29 million 897,000 every September 30th beginning 30th 2026 Bond payable (09/30/2026 installment) _________________ Bond payable (other than 09/30/2026 installment) ________________ Interest payable ___________________
The amounts due on December 31, 2026 are: Bond payable (09/30/2026 installment) = $29,897,000, Bond payable (other than 09/30/2026 installment) = $90,313,578.19,Interest payable = $622,578.19
We can break down the problem into two parts: first, calculating the amount of bonds payable that are due on September 30, 2026 (the first installment), and second, calculating the amount of bonds payable and interest payable that are due on any other date.
Bond payable due on September 30, 2026:
The total amount of bonds payable outstanding is $119,588,000. The installment due on September 30, 2026 is $29,897,000. Therefore, the amount of bonds payable that is due on September 30, 2026 is:
$29,897,000
Bond payable and interest payable due on any other date:
The total amount of bonds payable outstanding is $119,588,000. The first installment of $29,897,000 is due on September 30, 2026, which leaves $89,691,000 of bonds payable outstanding.
The bonds pay 12% interest every September 30th, so we need to calculate the interest payable for the period from September 30, 2026 to any other date. Let's assume we want to calculate the bond payable and interest payable due on December 31, 2026 (the end of the year).
The period from September 30, 2026 to December 31, 2026 is 92 days (or 3 months). The annual interest rate is 12%, so the daily interest rate is:
12% / 365 = 0.0328767%
The interest payable for 92 days is therefore:
$89,691,000 x 0.0328767% x 92/365 = $622,578.19
Adding this to the outstanding bonds payable gives:
Bond payable (other than 09/30/2026 installment) = $89,691,000 + $622,578.19 = $90,313,578.19
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What is an equation of the line that passes through the points (−2,4) and (3,−6)?
Answer:
y=10x+24
Step-by-step explanation:
y-4=10(x-(-2))
y-4=10(x+2)
y-4=10x+20
y=10x+24
Answer:
y= -2x
Step-by-step explanation:
m=y2-y1/x2-x1
m=-6-4/3-(-2)
m=-2
y= mx+q
y= -2x+q
-6= -2(3)+q
-6+6= q
q= 0
Therefore y= -2x
2x +5y +1=0
3x + 7y= 1
Answer:
We can solve this system of equations using substitution or elimination.
Using substitution, we can solve for one variable in terms of the other in one equation and substitute that expression into the other equation. For example, we can solve the first equation for x in terms of y:
2x + 5y + 1 = 0
2x = -5y - 1
x = (-5/2)y - 1/2
Now we can substitute this expression for x into the second equation:
3x + 7y = 1
3((-5/2)y - 1/2) + 7y = 1
-15/2 y - 3/2 + 7y = 1
9.5y = 5
y = 5/9.5 = 1/1.9
Now we can substitute this value of y back into either equation to solve for x:
x = (-5/2)y - 1/2
x = (-5/2)(1/1.9) - 1/2
x = -2.632
So the solution to the system of equations is x = -2.632 and y = 1/1.9.
Write the quadratic equation y=x^2-6x-16
Use the information given to answer the question
A pack of crayons costs 1. 25. If z packs of crayons are purchased, the total price, P, can be represented by the relation P = 1. 252.
Does P 1. 252 represent a function, and why?
The relation is a function because there is exactly 1 value of P for every value of 1.
The relation is a function because there is exactly 1. 25 value of P for every value of z.
The relation is not a function because there is exactly 1. 25 value of P for every value of 2.
The relation is not a function because there is not exactly 1 value of P for every value of z.
Sting
The right response is: Because there is precisely one value of P for every value of z, the relationship is a function. The value that a function produces after accepting one or more input values is known as the output value of the function.
The right response is:
Because there is precisely one value of P for every value of z, the relationship is a function.
Each input value (in this case, z) in a function has a single related output value (P in this case). This requirement is met by the relation stated since there is exactly one value of P for each value of z.
The output value of a function given an input value x is denoted as f(x) in mathematical notation. This is the value that is obtained after applying the function to the input value.
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Which one of the following graphs represents correct represent the variation of the gravitational field (E) with the distance (r) from the centre of a spherical shell of mass M radius R?
Option D is the graph that represent the variation of the gravitational energy (E) with the distance (r) from the centre of a spherical shell of mass M radius R
How to solve for the graphWhen R is less than r, the graph we would have is a straight line. When R = r, it has constant value and finally if R is more than r it is a hyperbola.
The intensity that is inside the spherical shell would be given as 0. Then we would have this represented as
I = 0
upto r = R
I ∝ I/r²
r > R
Hence D
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An entomologist has an insect colony for an experiment. The
population of insects is increasing at a
continuous rate of 5.6 percent per week. The
initial population of the insect colony was 100.
The function that models the insect population after tt weeks is
P(t)=
Using the function from part (a) we can estimate that the insect population after 16 weeks is
___
(round your answer to the nearest whole number)
After how many weeks will the population reach 3800 insects?
weeks
___
(round your answer to one decimal place)
16 weeks is 277
3800 insects in about 27.4 weeks
The answer to the question is given below.An entomologist has an insect colony for an experiment. The population of insects is increasing at a continuous rate of 5.6 percent per week. The initial population of the insect colony was 100.The formula that represents the insect colony population after tt weeks is P(t) = 100 (1 + 0.056)^tUsing the formula from section (a), we can calculate that the insect population after 16 weeks is 277. The nearest whole number is 277.After how many weeks would the population of insects reach 3800?Let’s apply the formula P(t) = 100 (1 + 0.056)^t3800 = 100 (1 + 0.056)^t(1 + 0.056)^t = 38(1.056)^t = 38log 1.056^t = log 38t log 1.056 = log 38t = log 38 / log 1.056t ≈ 27.4 weeksThe insect population will reach 3800 insects in about 27.4 weeks. We rounded the answer to one decimal place.
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find W (hypotenuse) if the angle is 15 degrees and the adjacent side has a length of 7 cm
Answer:72
Step-by-step explanation:
71+1=72
Draw a line representing the "rise" and a line representing the "run" of the line. State
the slope of the line in simplest form.
Answer:yo
Step-by-step explanation:
Use a Double- or Half-Angle Formula to solve the equation in the interval \( [0,2 \pi \) ). (Enter your answers as a comma-separated list.) \[ \cos (\theta)-\sin (\theta)=\sqrt{2} \cdot \sin \left(\fr
The solution for the given equation isθ = π/2, 5π/6, 13π/6.
The given equation is cos(θ) - sin(θ) = √2sin(θ/2)cos(θ/2).Use double-angle formula for sin to write sin(θ) in terms of cosθ as follows:sin(θ) = 2sin(θ/2)cos(θ/2).Substitute sin(θ) with the above equation. We have,cos(θ) - 2sin(θ/2)cos(θ/2) = √2sin(θ/2)cos(θ/2)cos(θ) - 3sin(θ/2)cos(θ/2) = 0(cosθ - 3sin(θ/2) = 0 or cosθ/3 = sin(θ/2)Squaring both sides of the equation gives us:cos2θ/9 = 1 - cos2(θ/2)cos2(θ/2) = (1 - cos2θ/9) / 2cos2(θ/2) = (9 - cos2θ) / 18sin2(θ/2) = 1 - cos2(θ/2)sin2(θ/2) = 1 - ((9 - cos2θ) / 18)sin2(θ/2) = (9 + cos2θ) / 18Therefore, the solution for the given equation isθ = π/2, 5π/6, 13π/6.
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What are the solutions to this quadratic equation?
The solutions can be written as x = 8 + 10i and x = 8 - 10i
Describe Quadratic Equation?The graph of a quadratic equation is a parabola, which is a symmetrical U-shaped curve. The direction and shape of the parabola depend on the value of a. If a is positive, the parabola opens upward, and if a is negative, the parabola opens downward.
Quadratic equations are used in many areas of mathematics and science, including physics, engineering, and finance. They are also used in real-life situations to model and solve problems related to motion, distance, time, and other quantities.
To solve this quadratic equation, we can rearrange it to be in standard form:
x² - 16x + 164 = 0
We can then use the quadratic formula to find the solutions:
x = (-b ± √(b² - 4ac)) / 2a
In this case, a = 1, b = -16, and c = 164. Substituting these values into the formula, we get:
x = (16 ± √((-16)² - 4(1)(164))) / 2(1)
Simplifying, we get:
x = (16 ± √(256 - 656)) / 2
x = (16 ± √(-400)) / 2
We have a negative number inside the square root, which means that the solutions are complex. To express them in the form x = a + bi and x = a - bi, we can simplify the square root by factoring out -1:
x = (16 ± √(400) i) / 2
x = (16 ± 20i) / 2
x = 8 ± 10i
Therefore, the solutions to the quadratic equation x² - 16x + 164 = 0 are:
x = 8 + 10i and x = 8 - 10i
So, a = 8 and b = 10. Therefore, the solutions can be written as:
x = 8 + 10i and x = 8 - 10i
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What advice would you give to students B and C to help them avoid factoring this
type of problem incorrectly in the future?
Correct solved problem
Student A factored
this expression correctly:
²-10x-24
(x-12)(x+2)
Factor: ²-10-24
Incorrect solved problem:
Sum of the integers does not
equal the middle term
Student B did not factor
this expression correctly.
²-10x-24
(x-4)(x+6)
Incorrect solved problem:
Sum of the integers does not
equal the middle term
Student C did not factor
this expression correctly:
-10-24
(r+12)(x-2)
**Day 4 will open once you post to the discussion board
Therefore , the solution of the given problem of expressions comes out to be students can develop their factoring abilities and avoid mistakes with practise and instruction.
What exactly is an expression?Estimates that combine joining, disabling, and rather than randomly divide should be produced when variables are shifting. If they got together, they could solve a mental puzzle, provide some data, and instead software. A declaration of truth contains formulas, components, and mathematical processes like combination, subtraction, omission, and grouping. Both phrases and words can be assessed and analysed.
Here,
are some pointers to assist students B and C prevent incorrect factoring in the future:
Regularly practise factoring: Since factoring is a talent that must be honed, students should make sure to factor problems frequently. They will be better able to understand the procedure and spot trends in the expressions they are factoring as a result of this.
Verify the indications again because they are prone to error when factoring. Students should double-check their distribution of the signs and make sure no negative indicators are being left out.
Students should double-check their work by multiplying the factors back together after factoring an equation. They'll be able to correct any errors they may have made thanks to this.
If students need assistance with factoring, they should ask an instructor or tutor for assistance. Although factoring can be a challenging ability to master, students can develop their factoring abilities and avoid mistakes with practise and instruction.
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15 miles in 6 hours what is the average speed
15 miles in 6 hours was covered, hence with the average speed formula we find that the average speed is 2.5 miles per hour.
To find the average speed, we can use the formula:
Average Speed = Total Distance / Total Time
In this case, the total distance covered is 15 miles, and the total time taken is 6 hours. Substituting these values into the formula, we get:
Average Speed = 15 miles / 6 hours
Average Speed = 2.5 miles per hour
Therefore, the average speed for the 15 miles covered in 6 hours is 2.5 miles per hour. Average speed is a measure of how fast an object is moving over a certain distance during a certain time period. It is calculated as the ratio of the total distance traveled by an object to the total time it took to cover that distance. The unit of measurement for average speed is distance divided by time, such as miles per hour (mph), kilometers per hour (km/h), or meters per second (m/s). Average speed can be calculated for various types of objects in motion, including vehicles, runners, and even animals. It is often used to describe the efficiency of movement, and can be useful in making predictions about how long it will take to cover a certain distance or complete a certain task
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∠F and ∠U included sides and included angles?
given is ΔFUN
A triangle has three straight sides that meet at a point. The sides' lengths are adjustable, however the longest side's length cannot be longer than the sum of the lengths of the other two sides.
What method can be used to locate the triangle?A triangle has three straight sides that meet at a point. The sides' lengths are adjustable, however the longest side's length cannot be longer than the sum of the lengths of the other two sides.
In the triangle ΔFUN,
∠F is an angle formed by the vertices F, U, and N.
∠U is an angle formed by the vertices F, U, and N.
The included sides of ∠F and ∠U are:
The included side of ∠F is UN.
The included side of ∠U is FN.
The included angles of ∠F and ∠U are:
Therefore, The included angle of ∠ F is ∠FUN, which is formed by the sides FU and UN. The included angle of ∠U is ∠NUF, which is formed by the sides NU and UF.
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What is 52s+30b it is on an I ready
Another representation of the given algebraic expression 52s + 30b as required is; 2 (26s + 15b).
Which expression is equivalent to the given expression 52s + 30b?As evident in the task content; it follows that the equivalent expression to 52s + 30b is to be determined.
Given; 52s + 30b
Since 2 is a common factor to both terms of the equation; we have that;
2 (26s + 15b)
Ultimately, the equivalent expression is; 2 (26s + 15b).
Complete question:
What is 52s+30b in a factorized form.
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last question...this is geometry. 15 points to whoever gets it right and can explain
I need help I tried everything
John invest R500, at 12% per annum using simple interest for 4 years. How much money will John have at the end of 4 years?
Answer: 42$
Step-by-step explanation:
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Montrose Oil and Lube is a car care center specializing in ten-minute oil changes. Montrose Oil and Lube has two service bays, which limits its capacity to 4,000 oil changes per month. The following information was collected over the past six months: (Click the icon to view the information.) Read the requirements. 1. Prepare a graphing the volume of oil changes (x-axis) against the company's monthly operating expenses (y-axis). (Enlarge the graph to maximum size and use the point tool button displayed below to draw the graph.) Requirements 1. Prepare a scatterplot graphing the volume of oil changes (x-axis) against the company's monthly operating expenses (y-axis). 2. How strong of a relationship does there appear to be between the company's operating expenses and the number of oil changes performed each month? Explain. Do there appear to be any outliers in the data? Explain. 3. Based on the scatterplot, do the company's operating costs appear to be fixed, variable, or mixed? Explain how you can tell. 4. Would you feel comfortable using this information to project operating costs for a volume of 3,800 oil changes per month? Explain. Data table
To answer the question, let's start by preparing a scatterplot graphing the volume of oil changes (x-axis) against the company's monthly operating expenses (y-axis). From the data provided, the scatterplot appears as follows:
It appears that there is a fairly strong correlation between the number of oil changes performed and the company's operating expenses each month. The points appear to fall along a linear line, and there do not appear to be any outliers. This indicates that the company's operating costs are fixed.
Given the strong correlation between the number of oil changes and the company's operating expenses, we can feel confident using the data to project operating costs for a volume of 3,800 oil changes per month.
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In which number does the digit 2 have a value that is 1/10 times as great as the digit 2 in the number 6,257.11? Choose 1 answer:
(Choice A)
5,619.25
(Choice B)
291.93
(Choice C)
7,326.64
(Choice D)
2,575
In the number 7,326.64, the digit 2 appears in the thousands place, and its value is 200. Therefore, the answer is (Choice C) 7,326.64.
What is number?A number is a mathematical concept used to represent a quantity or value. Numbers can be used to count, measure, or calculate, and they are fundamental to many fields of mathematics, science, and everyday life. Numbers can be represented and manipulated using mathematical symbols, such as digits, operations, and functions. They are used in many applications, such as in finance, engineering, physics, and computer science.
Here,
To solve this problem, we need to find a number that has a digit 2 with a value that is 1/10 times as great as the digit 2 in the number 6,257.11.
In the number 6,257.11, the digit 2 appears in the thousands place, and its value is 2,000.
To find a number in which the digit 2 has a value that is 1/10 times as great, we need to divide 2,000 by 10, which gives us 200.
So, we need to look for a number in which the digit 2 appears in the thousands place and has a value of 200. The only option that satisfies this condition is (Choice C) 7,326.64.
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