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3 years ago

11

Write a real world problem in which you need to find the volume of a right rectangular prism. Solve your problem.

1 answer:

sveticcg [70]3 years ago

5 0

Do not trust people that share links these people hack and put viruses into your phone/cumputers

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Simplify:0.3(x-2y)+0.5x-y

astraxan [27]

Answer:

0.3[x-2y] + 0.5x - y

Use the distributive property: a(b-c) = ab - ac

0.3x - 0.6y + 0.5x - y

Now, add/subtract all like terms

0.3x + 0.5x - 0.6y - y

0.8x - 1.6y

If you want, we can simplify that down further to:

0.8(x - 2y)

7 0

3 years ago

Read 2 more answers

HELP!!!<br> y + 2 = 4/3(x - 7)

11111nata11111 [884]

Well I don't know if you are looking for the variable but here it is :)

7 0

4 years ago

What is the length of XPY in terms of pi?

rosijanka [135]

Answer:

i dont know

Step-by-step explanation:

i don know

8 0

4 years ago

If f(x) = 3x – 12, what is f(2)?

fredd [130]

It would be B. -6.

f(x) = 3x-12
f(2) = 3(2)-12
f(2) = 6-12
f(2) = -6

5 0

3 years ago

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Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

4 years ago

Read 2 more answers