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Allisa [31]
3 years ago
5

y varies jointly as x and the cube of z; y=432 when x= 4 and z=3. find the constant variation of k and the variation equation

Mathematics
1 answer:
polet [3.4K]3 years ago
5 0

Answer: constant of variation k = 36 the variation equation, y = 36xz^3

Step-by-step explanation:

y varies jointly as x and the cube of z. This means that y varies directly as x and directly as z^3

In order to remove the "variation" symbol and replace it with "equal to" symbol, we will introduce a constant of variation, k

y = kxz^3

y=432 when x= 4 and z=3

We will substitute these values of x, y and z into the equation to determine the value of k

432 = k × 4 × 3

432 = 12k

k = 432/12 = 36

We will substitute the value of k into the equation. Substituting k = 36 into equation y = kxz^3

The equation becomes

y = 36xz^3

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I need the answer and show work for this question
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Answer:

f(3) =-12

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Step-by-step explanation:

HELLO THERE

f(3)=-2(3)^2+8(3)

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Image a coordinate plane with points plotted at ordered pairs Q(-4,4) R(2,4) S(5,-3) T(2,-3) and U (-4,-3) what is the area of t
pogonyaev

In counterclockwise order that's


R(2,4), Q(-4,4), U(-4,3), T(2,-3), S(5,-3)


That's a trapezoid, height 7 upper base 6, lower base 9, so average base 15/2 and area 7(15)/2 = 105/2 = 52.5.


Answer: 52.5


The shoelace theorem gives the answer without thinking. We write the list on top of itself shifted and add up the cross produces:


R(2,4), Q(-4,4), U(-4,-3), T(2,-3), S(5,-3)

Q(-4,4), U(-4,-3), T(2,-3), S(5,-3), R(2,4),


A = \frac 1 2 | 2(4)-(4)(-4) + -4(-3)-(-4)(4) + -4(-3)- -3(2) + 2(-3)-(-3)(5) + 5(4)- -3(2)| = \frac{105}{2} \quad\checkmark




8 0
3 years ago
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