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2 years ago

Question is attached will give brainliest ty !!

Mathematics

1 answer:

creativ13 [48]2 years ago

6 0

Answer:

A. (2x, 2y)

Step-by-step explanation:

Original figure: PQRTS

Dilated figure: P'Q'R'S'T'

Take one of the corresponding sides of the figure and find the scale factor:

SR = 2, S'R' = 4
Scale factor is 4/2 = 2

The dilation rule as per above is:

(x, y) → (2x, 2y)

Correct option is A

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The number of men in a tour group is 15 times the number of women. If there are 48 people in the group, how many are men? A. 45

ankoles [38]

A. 45 hope this is the correct answer if not, sorry!

3 0

3 years ago

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2) A cake in the shape of a circus tent is used as a centerpiece at a celebration. The cake consists of a cylinder and a cone.Th

zavuch27 [327]

You have:

- The diameter of the cylinder is 12 inches and its height is 14 inches.
-The height of the cone is 6 inches.

So, you must apply the formula for calculate the volume of the cylinder a the formula for calculate the volume of a cone.

V1=πr²h

V1 is the volume of the cylinder.
r is the radius.
h is the height (h=14 inches)

The problem gives you the diameter, but you need the radius, so you have:

r=D/2
r=12 inches/2
r=6 inches

When you substitute the values into the formula, you obtain:

V1==πr²h
V1=(3.14)(6 inches)²(14 inches)
V1=1582.56 inches³

The volume of the cone is:

V2=(πr²h)/3

V2 is the volume of the cone.
r is the radius (r=6 inches)
h is the height of the cone (h=6 inches).

Then, you have:

V2=(πr²h)/3
V2=(3.14)(6 inches)²(6 inches)/3
V2=226.08 inches³

Therefore, the volume of the cake (Vt) is:

Vt=V1+V2
Vt=1582.56 inches³+226.08 inches³
 Vt=1808.6 inches³

8 0

3 years ago

Write and solve an equation to find the value of X

Mumz [18]

Angles on a straight line add up to 180 degrees so

45 + 3x = 180
Subtract 45 from both sides to remove the 45 on the left

3x = 135
Then divide both sides by 3
x = 45

Answer is X = 45

8 0

3 years ago

Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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2 years ago

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Norma-Jean [14]

Answer:

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2 years ago