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dimulka [17.4K]
2 years ago
8

Need help on geomrty homework

Mathematics
1 answer:
Gnesinka [82]2 years ago
8 0

Answer:

if I am in Virginia, then I am in Richmond

Step-by-step explanation:

this is the correct way of the statement

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30= 2w + 2h<br><br> W= 3/4 h
Alexeev081 [22]

Answer:

h=60/7

w=45/7

Step-by-step explanation:

just try it.

7 0
3 years ago
Read 2 more answers
7/18 is closes to?<br><br>1/2<br><br>0<br><br>1
Anika [276]

Answer:

1/2

Step-by-step explanation:

Just divide 7 by 18.

7/18 = 0.38888888889

6 0
3 years ago
Please do both please answer properly and show how u got ur answer
Sphinxa [80]

Answer:

a) No.

b) Yes.

Step-by-step explanation:

a) It's not proportional. Although the graph represents a straight line, it does not pass through the origin (0,0).

b) It is proportional. It is a straight line that passes through the origin.

6 0
2 years ago
A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len
aniked [119]

Answer:

a. V = (20-x) x^{2} in^{3}  

b . 1185.185 in^{3}

Step-by-step explanation:

Given that:

  • The height:  20  - x (in )
  • Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h *x^{2} in^{3}

<=> V = (20-x) x^{2}  in^{3}

b. What is the maximum possible volume of the box?

To  maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3x^{2} + 40x

Let dV / Dx = 0, we have:

-3x^{2} + 40x  = 0

<=> x = 40/3

=>the height h = 20/3

So  the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 in^{3}

7 0
2 years ago
Consider the initial value problem y′′+36y=2cos(6t),y(0)=0,y′(0)=0. y″+36y=2cos⁡(6t),y(0)=0,y′(0)=0. Take the Laplace transform
Bingel [31]

Recall the Laplace transform of a second-order derivative,

L(y''(t)) = s^2Y(s)-sy(0)-y'(0)

and the transform of cosine,

L(\cos(at))=\dfrac s{a^2+s^2}

Here, both y(0)=y'(0)=0, so taking the transform of both sides of

y''(t)+36y(t)=2\cos(6t)

gives

s^2Y(s)+36Y(s)=\dfrac{2s}{36+s^2}

\implies Y(s)=\dfrac{2s}{(s^2+36)^2}

4 0
3 years ago
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