Help please[tex]\sqrt{x-3} -\sqrt{x} =3[/tex]

The linear function is C(x) = 500 + 1.25x and the revenue function is R(x) = 3.50x.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

From the data:

The linear function:

C(x) = 500 + 1.25x

For revenue:

R(x) = 3.50x

R(50) = 3.50(50) = $175

Using the C(x) and revenue function we can find the number of slushies and profit.

Thus, the linear function is C(x) = 500 + 1.25x and the revenue function is R(x) = 3.50x.

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They each will pass 12 scales.

What do you mean by algebraic expression?

The concept of algebraic expressions is the use of letters or alphabets to represent numbers without providing their precise values. We learned how to express an unknown value using letters like x, y, and z in the fundamentals of algebra. Here, we refer to these letters as variables. Variables and constants can both be used in an algebraic expression. A coefficient is any value that is added before a variable and then multiplied by it.

It is given that Becky has passed 4 scales, whereas her friend Charlotte has passed 8 of them.

Going forward, Becky has committed to passing 4 scales per week, and Charlotte has committed to passing 2 per week.

Let number of weeks be x

Becky has passed 4x + 4

Charlotte has passed 2x + 8

According to question:

4x + 4 = 2x + 8

4x - 2x = 8 - 4

2x = 4

x = 4/2 = 2

Becky has passed 4x + 4 = 4(2) + 4 = 8 + 4 = 12

Charlotte has passed 2x + 8 = 2(2) + 8 = 4 + 8 = 12

Therefore, They each will pass 12 scales.

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[tex]\begin{array}{llll} ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} \end{array} ~\hfill \begin{array}{llll} \textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \end{array} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{6x^{-1}y^{-1}}{\sqrt[4]{xy^4}}\implies \cfrac{6x^{-1}y^{-1}}{\left( xy^4 \right)^{\frac{1}{4}} }\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^{4\cdot \frac{1}{4}}}\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^1}\implies \cfrac{6x^{-1}y^{-1}x^{-\frac{1}{4}} y^{-1}}{1}[/tex]

[tex]6x^{-1}x^{-\frac{1}{4}}y^{-1} y^{-1}\implies 6x^{-1-\frac{1}{4}} y^{-1-1}\implies {\Large \begin{array}{llll} \underset{\textit{\small a}}{6} \stackrel{\textit{\small ~~ b}}{x^{-\frac{5}{4}}} \stackrel{\textit{\small ~~ c}}{y^{-2}} \end{array}}[/tex]

Answer:

Slope- y=1/4x-6

Y intercept= (0,-6)

Hope this helps!

Answer:

15d10mo

Step-by-step explanation:

Simplify:

(9+6)(mod10)

=15d10mo

Answer:

(1, -2)

Step-by-step explanation:

The line is tangent to the parabola, so it intersects in only 1 point.

Answer: (1, -2)

The berries is the bowl is 6 berries fewer than the original amount

The amount of berries in the bowel is solved using addition and subtraction property of equality

36 berries are in a bowl James eats 21 of the berries

= 35 berries - 21 berries

= 15 berries

he put 14 more berries in the bowl

= 15 berries + 14 berries

= 29 berries

How many fewer is solved by

= 35 berries - 29 berries

= 6 berries

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Answer:

The probability of a head of a die is 1/6 and that of a side of a coin

Answer: The answer is 2

Step-by-step explanation:

3j + 4 = 10  J = 2

3x2=6

6+4= 10

The perimeter of the image of the triangle is approximately equal to 87 units.

In this problem we find the coordinates of the vertices of a triangle: A(x, y) = (- 4, 3), B(x, y) = (4, 3), C(x, y) = (- 4, - 6), of which we must determine the perimeter of its image, which is the result of a scale factor of 3. The perimeter is the sum of the lengths of the triangle. The length of each side of the image of the triangle can be found by Pythagorean theorem and definition of dilation:

l = √[(Δx)² + (Δy)²]

Where:

And the perimeter of the image of the triangle is described below:

p = k · (AB + BC + AC)

Where k is the scale factor.

First, determine the lengths of sides AB, BC and AC:

Side AB

AB = √[[4 - (- 4)]² + (3 - 3)²]

AB = 8

Side BC

BC = √[(- 4 - 4)² + (- 6 - 3)²]

BC = √[(- 8)² + (- 9)²]

BC = √145

Side AC

AC = √[(- 4 - 4)² + (- 6 - 3)²]

AC = 9

Second, determine the perimeter of the image:

p = 3 · (8 + √145 + 9)

p = 3 · (17 + √145)

p = 51 + 3√145

p ≈ 87.125

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Answer:

Option 2 —› (56 - 42) / 7 = 2

Step-by-step explanation:

First you use PEMDAS to check.

P- parentheses

E- Exponents

M- Multiplication (can be switched with D)

D- Division (can be switched for M)

A- Addition

S- Subtraction

Now you need to solve each problem.

Option 1: INCORRECT

(56 - 42) / 7 = 5

(14) / 7 = 5

2 = 5

Option 2: CORRECT

(56 - 42) / 7 = 2

(14) / 7 = 2

2 / 2

I'll do the rest, so you can see how to solve.

Option 3: INCORRECT

56 - (42 / 7) = 2

56 - 6 = 2

50 = 2

Option 4: INCORRECT

(56 - 7) / 42 = 5

(49) / 42 = 5

1 1/6 = 5

Answer:

A = 29°

Step-by-step explanation:

set

(3x-4) + (9x+17) + (6x-31) = 180

Solve for x

18x - 18= 180

18x = 198

x = 11

So A will be

A = 3x - 4

A = 3(11) - 4

A = 29°

Answer:

Step-by-step explanation:

Answer: 54

Step-by-step explanation: 3 x 30seconds = 90 seconds (1 1/2)

3 x 18 = 54

Answer: C. 2x+3

Step-by-step explanation:

[tex]\displaystyle\\V=\frac{\pi r^2h}{3} \\\\V=\frac{\pi r^2(6)}{3}\\\\V=2\pi r^2\\\\\\8\pi x^2+24\pi x+18\pi \\\\V=2\pi (4x^2+12\pi +9)\\\\V=2\pi ((2x)^2+2(2x)(3)+3^2)\\\\V=2\pi (2x+3)^2\\\\THUS,\\[/tex]

[tex]2\pi r^2=2\pi (2x+3)^2[/tex]

Divide both parts of the equation by 2π:

[tex]r^2=(2x+3)^2[/tex]

Extract the square root of both parts of the equation:

[tex]r=2x+3[/tex]

The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5.

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

In other words, an equation is a set of variables that are constrained through a situation or case.

Let's say that number is x,

Quotients of 20 and x will be given as 20/x

20/x - 4 = 0

20/x = 4

x = 20/4 = 5

Hence"The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5".

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The probability that the total showing is less than 10 is 5/6.

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. The formula to calculate the probability of occurrence of an event 'A' can be written as -  

P(A) = n(A)/n(S)

where -

n(A) = Number of outcomes favorable to event A.

n(S) = Total number of outcomes

Given are two dices rolled.

The total number of favorable outcomes can be calculated as -

[a] = (1,1) to (1,6) = 6

[b] = (2,1) to (2,6) = 6

[c] = (3,1) to (3,6) = 6

[d] = (4,1) to (4,5) = 5

[e] = (5,1) to (5,4) = 4

[f] = (6,1) to (6,3) = 3

n(A) = [a] + [b] + [c] + [d] + [e] + [f] =  6+6+6+5+4+3 = 30

n(S) = 36

So, the probability of occurrence of an event 'A' -

P(A) = n(A)/n(S)

P(A) = 30/36

P(A) = 5/6

Therefore, the probability that the total showing is less than 10 is 5/6.

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The side lengths of the triangle are 89 ft, 132 ft and 159 ft

The given parameters are

Angle, A = 34 degrees

Hypotenuse = 159 ft

The length BC is calculated as

sin(A) = BC/Hypotenuse

So, we have

BC = Hypotenuse * sin(A)

This gives

BC = 159 * sin(34 degrees)

Evaluate

BC = 89 ft

The length AC is calculated as

cos(A) = AC/Hypotenuse

So, we have

AC = Hypotenuse * cos(A)

This gives

AC = 159 * cos(34 degrees)

Evaluate

AC = 132 ft

Here, we have

Angle, A = 34 degrees

Because the triangle is a right triangle, we have

Angle, C = 90 degrees

The sum of angles in a triangle is 180

So, we have

Angle B = 180 - 90 - 34

Angle B = 56

This is calculated as

Area = 0.5 * BC * AC

So, we have

Area = 0.5 * 89 * 132

Evaluate

Area = 5874

Hence, the area is 5874 square units

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The answer is 10.67. It can be calculated by integrating and then applying the limits.

What is the integration of [tex]x^n[/tex]?

The integration of power of a variable is given by the following formula:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

If limits are there then limits can be applied after integration.

Consider the given function.

[tex]\int^3_1(x^2-4x+9)dx[/tex]

Now, apply integration on each term as follows:

[tex]\int^3_1x^2dx-4\int^3_1xdx+9\int^3_1dx[/tex]

Now, use the following formula to integrate:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

Now, the integration will look like as follows:

[tex][x^3/3]^3_1-4[x^2/2]^3_1+9[x/1]^3_1[/tex]

Now, apply limits to find the answer as follows:

[tex][3^3/3-1^3/3]-4[3^2/2-1^2/2]+9[3-1]\\[/tex]

[tex][9-\frac{1}{3}]-4[9/2-1/2]+18\\10.67[/tex]

Hence, the answer is 10.67. It can be calculated by integrating and then applying the limits.

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Answer:

k = 7

Step-by-step explanation:

given y varies jointly with x and w then the equation relating them is

y = kxw ← k is the constant of variation

to find k use the condition when y = - 42, x = 2 and w = - 3 , then

- 42 = k × 2 × - 3

- 42 = - 6k ( divide both sides by - 6 )

7 = k

constant of variation is k = 7

Part A: Probability that one of 400-page novels will have exactly one error-free page is 24.06%.

Part B.) No more than three pages contain errors: 40.06%.

The given data in the question-

The following is the probability mass function (PMF) for X:

P(X = x) = ⁿCₓ . pˣ . qⁿ⁻ˣ

Part A.) One of 400-page novels will have exactly one error-free page.

P(x = 1)  = ⁿCₓ . pˣ . qⁿ⁻ˣ

n = 400, x = 1

P(x = 1)  = ¹⁰⁰C₁ . (0.005)¹ . (0.995)¹⁰⁰ ⁻ ¹

P(x = 1)  =  0.2706

Thus,  probability that one of 400-page novels will have exactly one error-free page is 24.06%.

Part B.) No more than three pages contain errors.

P(x ≤ 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)

P(x = 0)  = ¹⁰⁰C₀ . (0.005)⁰ . (0.995)¹⁰⁰ ⁻ ⁰

P(x = 0)  = 0.01346

P(x = 1)  = ¹⁰⁰C₁ . (0.005)¹ . (0.995)¹⁰⁰ ⁻ ¹

P(x = 1)  = 0.0270

P(x = 2)  = ¹⁰⁰C₂ . (0.005)² . (0.995)¹⁰⁰ ⁻ ²

P(x = 2)  = 0.00068

P(x = 3)  = ¹⁰⁰C₃ . (0.005)³ . (0.995)¹⁰⁰ ⁻ ³

P(x = 3)  = 0.0000027

Thus, P(x ≤ 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)

P(x ≤ 3) =  0.01346 + 0.00068 + 0.00068 + 0.0000027

P(x ≤ 3) =  0.4060021

Consequently, the likelihood of at maximum three pages having typographical errors is 40.06%.

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The athletic director can therefore draw the conclusion that the average academic standing of football players is comparable to the average standing of other student athletes.

The following formula is used to determine the test statistic:

SRS, s = 0.54, n = 40 > 30,

x = 3.18

H0: μ = 3.10

H1: μ > 3.10

α = 0.05

The t-distribution table and our alpha level of.05 are then used to determine our crucial values.

Our critical values are 2.093 for a two specimens with 19 levels of freedom and α = 0.05 level of significance.

t = x -  μ / s/√n

t = 3.18 - 3.10/ 54/√20

t = 0.66

As a result, It is clear that sampling error is the only possible explanation for the discrepancy between both the sample mean as well as the hypothesized value.

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Answer:

7

Step-by-step explanation:

seven is the value e you would get

Answer:

Step-by-step explanation:

24

Answer: 24!

Step-by-step explanation:

hope this help