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3 years ago

Which series of transformations will ALWAYS form an image that is congruent to its pre-image? Circle all that apply.

Mathematics

1 answer:

Law Incorporation [45]3 years ago

4 0

I think c not quite sure

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Help on this math questions please!

padilas [110]

Answer:

$y = - 3 {x}^{2} ( {x}^{2} + 8x + 15) = \\ =- 3 {x}^{2} ( (x + 3)(x + 5)) \\ roots \\ 0 \\ - 3 \\ - 5$

6 0

3 years ago

What is the value of the expression below? 32^3/5 O A. 6/5 B. 6 O c. 96/5 O D. 8

Hoochie [10]

Answer:

D. 8

Step-by-step explanation:

32^(3/5) is the same as $\sqrt[5]{32}$ ^3.

$\sqrt[5]{32}$ = 2 because 2^5=32

2^3 = 8

6 0

2 years ago

Read 2 more answers

For dinner, Javier will choose one of four fish options and one of five side dishes. His choices for fish are

rusak2 [61]

Answer:

Step-by-step explanation:

Please don't report I tried my best to solve it

7 0

3 years ago

Work out the volume of the shape

pashok25 [27]

Answer:

$\large\boxed{V=\dfrac{1,421\pi}{3}\ cm^3}$

Step-by-step explanation:

We have the cone and the half-sphere.

The formula of a volume of a cone:

$V_c=\dfrac{1}{3}\pi r^2H$

r - radius

H - height

We have r = 7cm and H = (22-7)cm=15cm. Substitute:

$V_c=\dfrac{1}{3}\pi(7^2)(15)=\dfrac{1}{3}\pi(49)(15)=\dfrac{735\pi}{3}\ cm^3$

The formula of a volume of a sphere:

$V_s=\dfrac{4}{3}\pi R^3$

R - radius

Therefore the formula of a volume of a half-sphere:

$V_{hs}=\dfrac{1}{2}\cdot\dfrac{4}{3}\pi R^3=\dfrac{2}{3}\pi R^3$

We have R = 7cm. Substitute:

$V_{hs}=\dfrac{2}{3}\pi(7^3)=\dfrac{2}{3}\pi(343)=\dfrac{686\pi}{3}\ cm^3$

The volume of the given shape:

$V=V_c+V_{hs}$

Substitute:

$V=\dfrac{735\pi}{3}+\dfrac{686\pi}{3}=\dfrac{1,421\pi}{3}\ cm^3$

7 0

3 years ago

A rectangle measures sixteen centimetres by three centimetres what is its area

harkovskaia [24]

A = bh
b = 16
h = 3
A = 16*3
A = 48 square centimenters

5 0

3 years ago

Read 2 more answers