CAN YOU GUYS HELP ME

The area of the circle to the nearest tenth is 28.3 square centimeters.

To find the area of a circle with a given radius, we use the formula

A = π[tex]r^2,[/tex]

where A is the area and r is the radius.

In this case, the radius is 3 cm, so we can substitute it into the formula to get:

A = 3.14 x [tex]3^2[/tex]

Simplifying this equation, we get the following:

A = 3.14 x 9

A = 28.26

To round this to the nearest tenth, we look at the digit in the hundredth place, 6. Since 6 is greater than or equal to 5, we round up the number in the tenth place, which is 2. Therefore, the final answer is:

A ≈ 28.3 [tex]cm^2[/tex]

So, the area of the circle to the nearest tenth is 28.3 square centimeters.

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The bearing of B from C is calculated to be 115 degrees

The bearing of D from B is calculated to be 213.65 degrees

The bearing from B to C is worked using SOH CAH TOA

Sin = opposite / hypotenuse - SOH

Cos = adjacent / hypotenuse - CAH

Tan = opposite / adjacent - TOA

The direction of movements describes a right triangle of

opposite = 30 km

adjacent = 30 km

The bearing is calculated using tan, TOA let the angle be x

tan x = Opposite / Adjacent

tan x = 30 / 30

tan x = 1

x = arc tan 1

x = 45 degrees

The bearing is 90 + 45 = 115 degrees

The bearing of D from B

x = arc tan (45/30)

x = 56.31

90 - 56.31 = 33.65

180 + 33.65 = 213.65 degrees

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Check the picture below.

so we're really looking for the area of a sector of a circle with 63° and a radius of 3, twice.

[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360} ~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ \theta =63\\ r=3 \end{cases}\implies A=\cfrac{(63)\pi (3)^2}{360} \\\\\\ A=\cfrac{63\pi }{40}\implies \stackrel{\textit{now let's double that}}{2\cdot \cfrac{63\pi }{40}}\implies \cfrac{63\pi }{20}\implies 9.9~cm^2[/tex]

Answer:

66.67%

Step-by-step explanation:

There are two ways to solve this problem depending on which way you like. Percentages are based on a 0-100 system and thus you can start by dividing 100/90. This will give you an amount of 1 family. We are looking for 60 families, so multiply that value by 60 to get how much of a percentage 60 families is.

The other way is to divide 90 by 60, and then multiply that result by 100 to give you a percent.

Regardless of your preferred method, both answers are the same.

Answer: The dance class has more students

Step-by-step explanation:

3 divided by 10 = 0.3

2 divided by 3 = 0.66

Since 0.66 is more than 0.3, we can say that more students entered the dance class than the music class.

Answer:

5, 6, 8, 11, 15, 20

5+1=6+2=8+3=11+4=15+5=20

18, 20, 24, 30, 38, 48

18+2=20+4=24+6=30+8=38+10=48

55, 54, 51, 46, 39, 30, 19

55-1=54-3=51-5=46-7=39-9=30-11=19

82, 81, 78, 73, 66

82-1=81-3=78-5=73-7=66

0+6=6+0

6=6

14+9=9+14

23=23

20x(4+3)=(20x4)+(20x3)

20x7=80+60

140=140

9+(6+5)=(9+6)+5

9+11=15+5

20=20

10x(4+6)=(10x8)+(10x2)

100=100

Step-by-step explanation:

The distribution of X is P(X = x) = 1/7 for x = 1, 2, 5, 10, 20, 50, 100 and Distribution of Y is P(Y = y) = 1/6 for y = 1, 2, 5, 10, 20, 50, 100, y ≠ X and their joint distribution is P(X = x, Y = x) = 0.

(a) The possible values of X and Y are as follows:

(i) X can be any of the 7 banknotes.

(ii) Y can also be any of the 7 banknotes, except the one that was already picked for X.

Since each of the 7 banknotes is equally likely to be picked, the probability distributions of X and Y are both uniform distributions over the set {1, 2, 5, 10, 20, 50, 100}:

P(X = x) = 1/7 for x = 1, 2, 5, 10, 20, 50, 100

P(Y = y) = 1/6 for y = 1, 2, 5, 10, 20, 50, 100, y ≠ X

The joint distribution of X and Y can be computed as follows:

P(X = x, Y = y) = P(X = x) × P(Y = y|X = x)

Since the second banknote must be smaller than the first, we have:

P(Y = y|X = x) = 1/(6 - 1) = 1/5, for y ≠ x

And for y = x:

P(Y = x|X = x) = 0

Therefore:

P(X = x, Y = y) = 1/7 × 1/5 = 1/35, for x ≠ y

P(X = x, Y = x) = 1/7 × 0 = 0

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The value of sin(D) is 7/25 after the application of the Pythagoras theorem.

The Pythagorean theorem is a fundamental theorem in geometry that describes the relationship between the sides of a right triangle. It claims that the hypotenuse's square length, which is the side that faces the right angle, is equivalent to the total of the squares of the lengths of the other two sides in a right triangle. The theorem can be formulated mathematically as:

c² = a² + b²

where, even the lengths for the remaining two sides (the legs) of the right triangle are a and b, and c is the length of the hypotenuse.

The Pythagorean theorem may be employed to determine the triangle's third side's length:

DE²= FD² + EF²

25² = 24² + EF²

625 = 576 + EF²

EF² = 49

EF = 7

Now, we can use the definition of sine to find sin(D):

sin(D) = opposite/hypotenuse = EF/DE = 7/25

Therefore, the value of sin(D) is 7/25.

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Answer:

see below.

Step-by-step explanation:

So first, find the perimeter.

94*2= 188

50*2=100

100+188=288ft  <----This is our perimeter

Next, find the length of the center line.

It is 50 feet.

Next, we are going to figure out the circumference of the big half circle.

C=[tex]\pi[/tex]d

C=(3.14)(44)=138.16 <------circumference of a WHOLE circle.

Divide 138.16 by 2 to get the half circle

138.16/2=69.08

There are 2 half circles, so multiply 69.08 by 2 to get 138.16

Next, we need to find the circumference of the center circle.

C=(3.14)(12)=37.68

Finally, add all the totals together

288+50+138.16+37.68=513.84 FEET <---dont forget units.

Answer:

0.099, 0.18, 0.2, 0.698.

Step-by-step explanation:

The solution of the equation -2 = c/16 by making c the subject of equation is c = -32

An equation is an expression containing numbers and variables linked together by mathematical operations such as addition, subtraction, division, multiplication and exponents.

Given the equation:

-2 = c/16

We are to solve for c by making c the subject of equation. To solve for c, we multiply both sides of the equation by 16 and simplify. Therefore:

-2 * 16 = c/16   * 16

Simplifying:

c = -32

The solution of the equation -2 = c/16 is c = -32

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The solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.

What is Linear Equation ?

Linear equation can be defined as equation in which highest degree is one.

We can solve for one variable using Equations #1 and #2, but not for all three variables. From Equation #2, we can solve for x in terms of y as x=5y+7. Substituting this expression for x into Equation #1 gives:

3(5y+7)+z+y=8

Simplifying this equation, we get:

16y+z=-13

We still have two unknowns, y and z, so we cannot solve for any variable using only Equations #1 and #2.

b. It is possible to solve for all three variables using Equations #1, #2, and #3. We can use the following steps to solve for the variables:

Use Equation #2 to solve for x in terms of y as x=5y-7.

Substitute this expression for x into Equation #3 to get:

3z+2(5y-7)-2y=15

Simplifying this equation, we get:

13y+3z=29

Substitute the expression for x from Step 1 into Equation #1 to get:

3(5y-7)+z+y=8

Simplifying this equation, we get:

16y+z=29

Solve for z in terms of y by subtracting the equation from Step 3 from the equation from Step 2:

(13y+3z)-(16y+z)=29-(-13)

Simplifying this equation, we get:

-3y+2z=42

Solve for z in terms of y by adding twice the equation from Step 1 to the equation from Step 4:

2(5y-7)+(-3y+2z)=42

Simplifying this equation, we get:

7y+2z=56

Solve for y by adding the equation from Step 4 to twice the equation from Step 5:

(7y+2z)+2(-3y+2z)=56+84

Simplifying this equation, we get:

5z=196

So, z=39.2. Substituting this value for z into the equation from Step 4 gives:

-3y+2(39.2)=42

Solving for y, we get y=-9.6. Finally, substituting the values for y and z into the equation from Step 1 gives:

x=5(-9.6)-7=-55.

Therefore, the solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.

c. To check if the solution found in part b holds for Equation #4, we substitute the values of x, y, and z into Equation #4 and see if the equation is satisfied:

4(-55)+5(-9.6)-2(39.2)=-3

Simplifying this equation, we get:

-220-48-78.4=-3

This equation is not satisfied, so the solution found in part b does not hold for Equation #4. Therefore, the system of equations does not have a unique solution.

Therefore, the solution for the system of equations using Equations #1, #2, and #3 is x=-55, y=-9.6, z=39.2.

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A decimal and fraction and shaded part of the first diagram second diagram is 100/100 or 1 and 0.63 or 63/100 respectively.

What is fraction?

A fraction is a way of expressing a part of a whole, or a part of a group, by using two numbers separated by a line. The number above the line is called the numerator, and the number below the line is called the denominator.

The first 10 by 10 matrix is fully colored, which means all 100 cells are colored. Therefore, the fraction of colored cells is:

100/100 = 1

The decimal representation of this fraction is simply 1.

For the second 10 by 10 matrix, the first 6 columns are fully colored, which means there are 6 x 10 = 60 colored cells. The 7th column has only the first 3 rows colored, which means there are 3 colored cells in this column. Therefore, the total number of colored cells is:

60 + 3 = 63

The fraction of colored cells is:

63/100

To convert this fraction to a decimal, we can divide the numerator by the denominator:

63 ÷ 100 = 0.63

Therefore, the shaded part of the second matrix is 0.63 or 63/100.

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The lateral surface area of the square is 784mm².

Having four equal sides, a square is a quadrilateral. There are numerous square-shaped objects in our immediate environment. Each square form may be recognised by its equal sides and 90° inner angles. A square is a closed form with four equal sides and interior angles that are both 90 degrees. Numerous different qualities can be found in a square.

If the square has a side length of 14 mm, then all sides are equal in length.

The lateral surface area of a square is simply the perimeter multiplied by the height of the shape. Since a square has equal length and width, the perimeter of the square is simply 4 times its side length. Therefore, we have:

Perimeter = 4 × 14 mm = 56 mm

The height of the shape is the same as the length of any of its sides, which is also 14 mm. Therefore, we have:

Lateral Surface Area = Perimeter × Height = 56 mm × 14 mm = 784 mm^2

Therefore, the lateral surface area of the square is 784mm².

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find the lateral surface area of a square box with dimensions 14 mm, 13mm and 6 mm.

The orbits of v and w imply a nonzero invariant subspace of dimension at most 2. All 2-dimensional irreducible representations of S3 are isomorphic and can be determined by the character table.

To show that V contains a nonzero invariant subspace of dimension at most 2, we analyze the G-orbits of v and w. Since x and y are the generators for S3, we can see that xv = x(u + xu + xều) = x²u + xều = u and yw = y(u + yu) = u + y²u = u. This implies that the G-orbits of v and w are the same, so V contains a nonzero invariant subspace of dimension at most 2. To prove that all irreducible two-dimensional representations of G are isomorphic, we must show that they all have the same character table. To do this, we can use the fact that the character table is an invariant of a representation, meaning that it is the same for all isomorphic representations. Therefore, all irreducible two-dimensional representations of G have the same character table, which can be determined by examining the character table for S3.

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The value of the function notations are;

a) d(20) = 10

b) d(0) = 0

c) d(20) < d(30)

d) d(0) = 0

e) d(30) = d(40)

f) d(t) = 5, when t = 5

The notation f(x) defines a function named f. This is read as “y is a function of x.” The letter x represents the input value, or independent variable. The letter y is replaced by f(x) and represents the output value, or dependent variable.

a) d(20)

The value of distance (d) when the time is 20 is; 10

b) d(0)

The value of distance (d) when the time is 0 is; 0

c) d(20) = 10

d(30) = 20

Thus;

d(20) < d(30)

d) The value of the time at d = 0 is 0. Thus;

d(0) = 0

e) d(30) = 20

d(40) = 20

Thus; d(30) = d(40)

f) d(t) = 5

The value of t at d = 5 is;

t = 5

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Step-by-step explanation:

Week 1, you eat 10 out of 95

Week 1 r ( 1) = 95 - 10(1) =85

Week 5 r(5) = 95 - 10r

=95 - 10(5)

=95 - 50

=45lbs

Week 8 r(8) = 95 - 10.r

=95 - 10(8)

=95 - 80

= 15 lbs

Week w r(w) = 95 - 10.w

= 95 - 10W (lbs)

f(r) = 95 - 10r

Where r is the number of weeks.

r(w) =35 means that at the end of w weeks, there were 35 candies left.

If the original candy was 95 and 35 was left after w weeks

r(w) = 95 - 10w

35 = 95 - 10w

Subract 95 from both sides

-60 = - 10w

Divide both sides by - 10

6= w

The solution means that 35 candies were left after w weeks and w represents 6weeks.

(you can c compare my answer with a second response)

The equation of the line that is parallel to y = -(1/4)x + 5 and passes through (2, -3) is y = -(1/4)x - 5/2.

To find the equation of the line that is parallel to the line y = −(1/4)x + 5 and passes through the point (2, −3), follow these steps:Step 1: Determine the slope of the given line.The slope-intercept form of the equation of the line is y = mx + b where m is the slope of the line. y = −(1/4)x + 5 is already in slope-intercept form, so its slope is −1/4. Step 2: Determine the slope of the line that is parallel to the given line. The slope of a line parallel to another line is the same as the slope of the given line. Therefore, the slope of the line we need to find is also −1/4. Step 3: Determine the y-intercept of the line we need to find. We already know that the line passes through the point (2, −3). To determine the y-intercept of the line, substitute x = 2 and y = −3 into the slope-intercept form of the equation of the line. −3 = −(1/4)(2) + b b = −3 + 1/2 = −5/2 Therefore, the y-intercept of the line we need to find is −5/2. Step 4: Write the equation of the line in slope-intercept form. The equation of the line we need to find is y = mx + b where m = −1/4 and b = −5/2. y = −(1/4)x − 5/2 is the equation of the line that is parallel to y = −(1/4)x + 5 and passes through (2, −3).Answer: The equation of the line that is parallel to y = -(1/4)x + 5 and passes through (2, -3) is y = -(1/4)x - 5/2.

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After 3 seconds, the rider is still at their starting height of 0 metres.

The height of a rider after 3 seconds can be calculated using the equation h(t) = v₀t - (1/2)gt², where h(t) is the height of the rider in metres at time t, v₀ is the initial velocity in metres per second, and g is the acceleration due to gravity in metres per second squared.

Assuming v₀ is 0, since the rider is just starting, and g is 9.81, then the height of the rider after 3 seconds is h(3) = 0 - (1/2)(9.81)(3²) = -44.355. Since this is a negative number, the rider is still at their starting height of 0 metres after 3 seconds.

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Answer:

We can solve the inequality for m to find the maximum number of muffins that can be purchased within the budget of $75:

75 ≥ 0.82m

Divide both sides by 0.82:

m ≤ 91.46

Since m must be a whole number, the maximum number of muffins that can be purchased is 91. Therefore, 91 customers could possibly get a muffin.

Step-by-step explanation:

The location at which a function is not differentiable is a Vertical tangent line

A function may not be differentiable at some points. Such points are known as non-differentiable points. Let's take a look at each of the given terms to figure out the non-differentiable points. Discontinuity: A discontinuity is when a function's graph is interrupted by a break or hole.

It occurs when a function is undefined at a certain point. It may be classified into three categories: removable, jump, and infinite. Functions may not be differentiable at removable discontinuities but are differentiable at jump and infinite discontinuities. A discontinuous point is not the same as a non-differentiable point because it may be differentiable at other points of the function.

Cusp: A cusp is a sharp corner formed by a curve. It happens when the slope of the function approaches infinity. The curve is not differentiable at the cusp. Horizontal Tangent Line: When the slope of a function approaches zero, it creates a horizontal tangent line. The function may or may not be differentiable at this point depending on the shape of the graph.

It may be differentiable or not differentiable. Therefore, it is not a non-differentiable point. Vertical Tangent Line: When a function's slope approaches infinity, it creates a vertical tangent line. The function is non-differentiable at this point. A vertical tangent line is always a non-differentiable point.

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Answer: -2x-10y

Step-by-step explanation:i'm in the 8th grade so i know im right

In the given cube, the required data is as follows:

Faces = 6

Edges = 12

Vertices = 8

A cube is a solid three-dimensional form with six square faces that all have the same length sides. It is one of the five platonic solids and is also referred to as a regular hexahedron.

Six square faces, eight vertices, and twelve edges make up the form.

Here in the question as asked,

Faces = 6

Edges = 12

Vertices = 8

Now to prove that we are correct,

V -E + F =2

= 8 - 12 + 6

=2

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If Secant Line RQ and tangent line QX, angle XRQ is labeled 56 degrees, then the measure of angle RXQ is 34 degrees.

Since the line QX is tangent to the circle R, we know that angle QXR is a right angle. We also know that angle XRQ is labeled as 56 degrees. Therefore, we can find angle RXQ using the fact that the sum of the angles in a triangle is 180 degrees.

We have:

m∠RXQ + m∠QXR + m∠XRQ = 180 degrees

Since m∠QXR is a right angle (90 degrees), we can substitute:

m∠RXQ + 90 degrees + 56 degrees = 180 degrees

Simplifying:

m∠RXQ = 34 degrees

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The logical expressions for each sentence can be written as follows:
(a) t(Sam, Math 101)
This expression states that Sam has taken Math 101.
(b) ∀x∃y t(x, y)
This expression states that for every student x, there exists a math class y such that the student x has taken the math class y.
(c) ∀x∃y (t(x, y) ∧ y ≠ Math 101)
This expression states that for every student x, there exists a class y such that the student x has taken the class y and the class y is not Math 101.
(d) ∃x∀y (t(x, y) ∧ y ≠ Math 101)
This expression states that there exists a student x such that for every math class y, the student x has taken the math class y and the math class y is not Math 101.
(e) ∀x∃y∃z (t(x, y) ∧ t(x, z) ∧ x ≠ Sam ∧ y ≠ z)
This expression states that for every student x, there exists two different math classes y and z such that the student x has taken the math classes y and z and the student x is not Sam.
(f) ∃y∃z (t(Sam, y) ∧ t(Sam, z) ∧ y ≠ z ∧ ∀w (t(Sam, w) → (w = y ∨ w = z)))
This expression states that there exists two different math classes y and z such that Sam has taken the math classes y and z and for every math class w, if Sam has taken the math class w, then the math class w is either y or z.

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The quotient using a synthetic method of division is m³ - 10m² - 9m - 1

The quotient expression is given as

(m⁴ -7m³ -39m² - 28m - 3) divided by m + 3

Using a synthetic method of quotient, we have the following set up

 -3 |   1    -7    -39    -28    -3

      |__________

Bring down the first coefficient, which is 1:

 -3 |   1    -7    -39    -28    -3

      |__________

         1

Multiply -3 by 1 to get -3, and write it below the next coefficient and repeat the process

 -3 |   1    -7    -39    -28    -3

     |____-3__ 30__27__3____

          1    -10    -9    -1      0

So, the quotient is m³ - 10m² - 9m - 1

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