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Fantom [35]
3 years ago
6

Please help what is the volume of the rectangular prism

Mathematics
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

120

Step-by-step explanation:

to find the volume of rectangular prisms, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units. After doing that, you get 120

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Suppose you have a right triangle with congruent legs and a hypotenuse that measure (12sqrt(5))/5 What is the length of the smal
anastassius [24]

The length of the smaller leg is 3.79

<h3>How to determine the length of the smaller leg?</h3>

Represent the smaller leg with x.

So, we have:

x^2 + x^2 = ((12\sqrt5)/5)^2 -- Pythagoras theorem

This gives

2x^2 = 144/5

Divide by 2

x^2 = 72/5

This gives

x^2 = 14.4

Take the square root

x = 3.79

Hence, the length of the smaller leg is 3.79

Read more about right triangles at

brainly.com/question/6322314

#SPJ1

4 0
2 years ago
Help please thanks!!!
Andreyy89
You already have the right Answer which is b
6 0
3 years ago
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I need to know what the variable is and please give me the equation.
Mumz [18]

Answer:

$9.50 = .25x

Step-by-step explanation:

38 quarters

6 0
2 years ago
How much is 7/9 of 3/14?
34kurt
7/9 of 3/14 is 1/6.   

Tell me if I'm wrong.  

Hope this helps :)

4 0
3 years ago
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Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
4 years ago
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