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3 years ago

15

HELP PLEASEE IM FAILING

2 answers:

Brrunno [24]3 years ago

8 0

Answer:

C

Step-by-step explanation:

∠ SVW = ∠ TWX = 84° ( corresponding angles are congruent )

∠ UVY = ∠ SVW = 84° ( vertical angles are congruent )

aleksklad [387]3 years ago

5 0

Answer:

failure is a step towards success!!

don't get urself drowned,,,,just get over it❤️

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A rectangular box has a volume x^3-64 cubic inches. If the height of the box x-4 inches, what is the area of the base of the box

Zanzabum

Answer:

$\purple{ \bold{ \therefore \: A_{base} = {x}^{2} + 4x + 16}}$

Step-by-step explanation:

$\because \: V_{box} = A_{base} \times height \\ \\ \therefore \: A_{base} = \frac{V_{box}}{height} \\ \\ \therefore \: A_{base} = \frac{ {x}^{3} - 64}{(x - 4)} \\ \\ \therefore \: A_{base} = \frac{ {x}^{3} - {4}^{3} }{(x - 4)} \\ \\ \therefore \: A_{base} = \frac{ \cancel{ ({x} - {4})}( {x}^{2} + 4x + {4}^{2} )}{\cancel{ ({x} - {4})}} \\ \\ \purple{ \bold{ \therefore \: A_{base} = {x}^{2} + 4x + 16}}$

8 0

3 years ago

PLEASE HELP! <br><br>Identify the y-intercept and zeros of <br>y = x^4 - 9x^2 - 36<br>Algebraically

const2013 [10]

Answer:

Our y-intercept is (0, -36).

Our real zeros are: (-2√3, 0) and (2√3, 0)

And our imaginary zeros are: (i√3, 0) and (-i√3, 0)

Step-by-step explanation:

We have the function:

$y=x^4-9x^2-36$

Let's identify the y- and x-intercepts.

Y-Intercept:

To determine the y-intercept, we will substitute 0 for x and solve for y. So:

$y=(0)^4-9(0)^2-36$

Evaluate:

$y=-36$

Therefore, our y-intercept is at (0, -36).

X-Intercepts:

To determine our x-intercepts, we need to substitute 0 for y and solve for x. So:

$0=x^4-9x^2-36$

Now, we can factor. Before doing so, we can make the factored easier to see by converting this into quadratic form. Let's let u=x². Then:

$0=(u^2)^2-9(x^2)-36$

Substitute:

$0=u^2-9u-36$

Now, let's factor. We can use -12 and 3. So:

$0=u^2+3u-12u-36 \\ 0=(u(u+3)-12(u+3)) \\ 0=(u-12)(u+3)$

Now, we can substitute back u:

$0=(x^2-12)(x^2+3)$

Zero Product Property:

$x^2-12=0\text{ or } x^2+3=0$

Solve for x in each case:

$x^2=12 \text{ or } x^2=-3$

Take the square root of both sides. Since we're taking an even root, we will need plus/minus. Therefore:

$x=\pm\sqrt{12}=\pm2\sqrt{3}\text{ or } x=\pm\sqrt{-3}=\pm i\sqrt{3}$

Therefore, our zeros (both real and imaginary) are: (2√3, 0), (-2√3, 0), (i√3, 0), and (-i√3, 0)

5 0

3 years ago

Find the volume of the planets

Natalija [7]

Answer:

can you show the planets. or do you want real planets volumes?

3 0

3 years ago

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Ratio equivalent to 24:56

ICE Princess25 [194]

3/7 is the final answer if you simplify is as much as possible.

4 0

3 years ago

What do you get when 1.293 and 150

mafiozo [28]

Answer:

450

Step-by-step explanation:

5 0

4 years ago