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bearhunter [10]
3 years ago
8

L*W*H is the formula for ___.

Mathematics
1 answer:
Vlad1618 [11]3 years ago
8 0
Volume of a rectangular prism.
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Find all the second partial derivatives. v = e5xey
Helga [31]
I'm taking the liberty of editing your  function  <span>v = e5xey:  It should be 
</span>
<span>v = e^5x^ey, with " ^ " indicating exponentiation.
</span>
Did you mean e^(5x) or (e^5)x?  I'll assume it's e^(5x).

The partial of   v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.

The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
5 0
3 years ago
Solve by any method <br> y= -7x+10<br> y= -3x+6
AnnyKZ [126]
Y=1
To get it, you use the method of substitution. So you swtich the top y by the equation of the bottom=-3x+6=-7x+10. You first add 7x to each side=4x+6=10 then you minus 6 from each side=4x=4 which is 1
4 0
3 years ago
What is the area of the shaded portion of the circle?
SOVA2 [1]

Answer:

The first option is the correct one, the area of the shaded portion of the circle is

[/tex](5 \pi -11.6)ft^2[/tex]

Step-by-step explanation:

Let us first consider the triangle + the shadow.

The full area of the circle is the radius squared times pi, so

A=(5 ft)^2 \cdot \pi \\25 ft^2 \cdot \pi

Since \frac{72^{\circ}}{360^{\circ}}=\frac{1}{5}, the area of the triangle + the shaded area is one fifth of the area of the whole circle, thus

A_1=\frac{1}{5}25 ft^2 \cdot \pi\\ =5 ft^2 \cdot \pi

If we want to know the area of the shaded part of the circle, we must subtract the area of the triangle from A_1.

The area of the triangle is given by

A_{triangle}=\frac{1}{2}\cdot (2.9+2.9)ft \cdot 4 ft\\= 11.6 ft^2

Thus the area of the shaded portion of the circle is

A_1-A_{triangle}=5 \pi ft^2-11.6ft^2\\= (5 \pi -11.6)ft^2

3 0
3 years ago
Find the perimeter.
kiruha [24]

Answer:

Hope this helps you

Step-by-step explanation:

Thank

8 0
2 years ago
One of the roots of the equation x^2−6x+q=0 is 4.5. Find the other root and the value of the coefficient q.
andre [41]

Answer:

Step-by-step explanation:

hello : here is an solution

3 0
3 years ago
Read 2 more answers
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