In the diagram, how many angles are supplementary but not adjacent angles with angle 7?

zero
one
two
three

The answer is 2304pi

The volume of the lollipop having the radius of 12 mm is 7241.14 [tex]mm^{3}[/tex].

Volume is the capacity of a container to hold something in its capacity. Like a drum can hold water.

The radius is given as 12 mm. The lollipop is having shape of sphere. The formula of volume of sphere is 4/3π[tex]r^{3}[/tex].

Volume =4/3π[tex]r^{3}[/tex].

Volume of lollipop is 4/3π[tex]12^{3}[/tex]

=4/3*22/7*576

=(88*1728)/21

=152064/21

=7241.14[tex]mm^{3}[/tex].

Hence the volume of lollipop having radius of 12 mm is 7241.14[tex]mm^{3}[/tex].

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Answer:

2) MN correspond to QR .

Step-by-step explanation:

Given : The two triangles below are similar.

To find : Which pair are corresponding sides

Solution : We have given that Triangle LMN is similar to triangle PQR.

Then Sides

MN correspond to QR .

Therefore, 2) MN correspond to QR .

Answer:d

Step-by-step explanation:

Answer:

21

Step-by-step explanation:

a = 6-3 = 3

b = 5 -(-2)= 7

Area = a*b = 3*7 = 21

The area of rectangle ABCD which is graphed in the coordinate plane with vertices A(3, -2), B(6, -2), C(6, 5), and D(3, 5) is 21 unit².

Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,

[tex]A=a\times b[/tex]

Here, (a)is the length of the rectangle and (b) is the width of the rectangle.

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(3, -2), B(6, -2), C(6, 5), and D(3, 5).

The shortest distance(length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:

[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]

The distance from A to B is the length of the rectangle. The distance of line segment AB is,

[tex]W= \sqrt{(3-6)^2 + (-2-(-2))^2}\\W = \sqrt{(-3)^2 + (0)^2}\\W=\sqrt{9}\\W=3\rm\; units[/tex]

The distance from B to C is the width of the rectangle. The distance of line segment BC is,

[tex]L = \sqrt{(6-6)^2 + (-2-5)^2}\\L = \sqrt{(0)^2 + (-7)^2}\\L=\sqrt{49}\\L=7\rm\; units[/tex]

Thus, the area of rectangle ABCD is,

[tex]A=7\times3\\A=21\rm\; unit^2[/tex]

Hence, the area of rectangle ABCD which is graphed in the coordinate plane with vertices A(3, -2), B(6, -2), C(6, 5), and D(3, 5) is 21 unit².

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Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

To determine which car is traveling faster at the checkpoint, we need to compare their velocities. Car A has a positive, constant acceleration, which means its velocity is increasing. Car B, on the other hand, travels at a constant speed, which means its velocity remains the same.

Therefore, Car A is traveling faster at the checkpoint than Car B.

Final answer:

Greg started with 20 quarters. The equation for the number of quarters saved after x days is y = 20 + 3x. With Greg's savings rate of 3 quarters per day, he cannot have exactly 100 quarters on any day since 100 is not a multiple of 3 plus the 20 starting quarters.

Explanation:

To determine how many quarters Greg started with, let 's' be the number quarters Greg started with, and 'd' be the number of quarters he saves each day. After 2 days, he had a total of 26 quarters saved, and after 5 days, he had a total of 35 quarters saved. These can be written as two equations:

Subtracting the first equation from the second gives:

3d = 9

From which we find that d = 3 (Greg saves 3 quarters a day). Plugging this value back into the first equation:

s + 2(3) = 26

s + 6 = 26

s = 20

Greg started with 20 quarters.

B. The equation to model the number of quarters Greg has saved, y, after x days is:

y = 20 + 3x

C. With the savings rate of 3 quarters a day, Greg's total will always be a multiple of 3 plus the 20 he started with. 100 is not a multiple of 3, meaning he can never save exactly 100 quarters by the end of any given day since 100 minus the 20 quarters he started with leaves 80, which is not divisible by 3.

Final answer:

Greg started with 20 quarters and saves 3 quarters every day afterward. The equation to model the number of quarters saved after x days is y = 20 + 3x. It is impossible for Greg to have exactly 100 quarters saved by the end of any given day because 100 cannot be reached by adding multiples of 3 to the starting amount of 20 quarters.

Explanation:

To solve for the number of quarters Greg started with and the number of quarters he saves each day, we can set up a system of equations. Let's designate q as the number of quarters Greg started with and d as the number of quarters he saves each day. From the problem, we have two points of data: On day 2, Greg has 26 quarters, and on day 5, he has 35 quarters.

The equations representing these two data points are:

Subtracting the first equation from the second gives us:

3d = 9

Dividing both sides by 3 gives us:

d = 3

Now that we have the value for d, we can substitute it back into the first equation:

q + 2(3) = 26

q + 6 = 26

Subtracting 6 from both sides gives us:

q = 20

Greg started with 20 quarters.

For the equation to model the number of quarters saved, y, after x days, we have:

y = q + dx

This simplifies to:

y = 20 + 3x

To address part C, let's analyze the possibility of Greg having exactly 100 quarters saved. If we set y to 100 in the equation y = 20 + 3x and solve for x, we get:

100 = 20 + 3x

80 = 3x

x = 80/3

x ≈ 26.67

Since x must be an integer because Greg cannot save a fraction of a day, he cannot have exactly 100 quarters by the end of any given day. This is because 100 is not a multiple of 3 (the daily amount Greg saves) when starting from 20. Thus, it's impossible for Greg to have exactly 100 quarters saved by the end of any given day.

Final answer:

To find the total number of classrooms in the school, we multiply the number of buildings (9) by the number of classrooms in each building (30), resulting in a total of 270 classrooms.

Explanation:

The question asks how many classrooms are there in total if a school has nine buildings, each with 30 classrooms. To find the answer, we need to multiply the number of buildings by the number of classrooms in each building. This calculation is straightforward:

Multiply the number of buildings (9) by the number of classrooms in each building (30).

This gives us: 9 buildings x 30 classrooms/building = 270 classrooms in total.

Therefore, the school has a total of 270 classrooms.