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FinnZ [79.3K]
3 years ago
12

PLEASE ANSWER QUESTION NOW

Mathematics
2 answers:
Alisiya [41]3 years ago
8 0
There should be 12 pounds left
Slav-nsk [51]3 years ago
4 0

⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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10 is subtracted from 2 times a number
Elena-2011 [213]

Answer: the equation is 2n-10

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Unit rates can be helpful with creating a ratio table and plotting the equivalent ratios on a coordinate plane. Explain why. PLE
uysha [10]

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Unit rate is often a useful means for comparing ratios and their associated rates when measured in different units. The unit rate allows us to compare varying sizes of quantities by examining the number of units of one quantity per one unit of the second quantity. This value of the ratio is the unit rate.

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Studentka2010 [4]

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2 years ago
Can you help me with question 6 please
Licemer1 [7]
I'm pretty sure the answer is 24
4 0
3 years ago
A rectangular solid has width w, a length of 7 more than the width, and a height that is equivalent to 15 decreased by 3 times t
dexar [7]

Answer:

For a rectangular solid with:

width = w

length = l

height = h

The volume is equal to:

V = w*l*h

in this case we know that:

width = w.

"length of 7 more than the width"

l = w + 7.

"a height that is equivalent to 15 decreased by 3 times the width."

h = 15 - 3*w

Then the volume will be:

V = w*l*h = w*(w + 7)*(15 - 3w) = (w^2 + 7*w)*(15 - 3*w)

V =  ( -3*w^3 + 15*w^2 + 105*w - 21*w^2)

V = (-3*w^3 - 6*w^2 + 105*w)

Now, the maximum volume will be for the value of w such that:

V'(w) = 0.

and:

V''(W) < 0

Where:

dV/dw = V'(w).

dV'/dw = V''(w)

Then first we need to differentiate the equation for the volume.

V'(w) = dV/dw = ( 3*(-3*w^2) + 2*(-6*w) + 105)

V'(w) = -9*w^2 - 12*w + 105.

Then we need to find the solution for:

-9*w^2 - 12*w + 105 = 0.

We can use the Bhaskara formula, and we will get:

w = \frac{+12 +-\sqrt{(-12)^2 - 4*(-9)*105} }{2*-9}  = \frac{+12 +- 62.6}{-18}

Then the two solutions are:

w₁ = (+12 - 62.6)/(-18) = 2.81

w₂ = (+12 + 62.6)/(-18) = -15.5

But we can not have a negative width, so we can just discard the second solution.

Now let's check the second condition for the maximum, we must have:

V''(2.81) < 0.

V'' = dV'/dw = 2*(-9*w) - 12 = -18*w - 12

V''(2.81) = -18*2.81 - 12 = -62.58 < 0 .

Then the volume is maximized when w = 2.81, and the maximum volume will be:

V(2.81) =  (-3*(2.81)^3 - 6*(2.81)^2 + 105*2.81) = 180.1

 

8 0
3 years ago
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