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3 years ago

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the perimeter of square f is 2/3 less than the perimeter of square g . the side lengths of square f are each x meters , and the

sides of square g are each 1/4(x-2/3) meters. which expression also represents the side lengths of square g A 1/4x - 1/6 B 1/4f - 2/3 C 1/4f - 3/7 D 1/4x - 11/12

2 answers:

iogann1982 [59]3 years ago

4 0

GIVEN:

Sides of square g = 1/4(x – 2/3)

ANSWER:

On simplyfying given sides,

1/4(x – 2/3)

1/4x – 1/4 × 2/3
1/4x – 1/6

And all other options are mismatch.

So, <u>Correct choice</u> - [A] 1/4x – 1/6

horrorfan [7]3 years ago

3 0

Answer:

A. 1/4x - 1/6

Step-by-step explanation:

1/4(x - 2/3) =
1/4x - 1/4(2/3) =
1/4x - 1/6

Correct choice is A

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You have been asked to design a can shaped like right circular cylinder that can hold a volume of 432π-cm3. What dimensions of t

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Answer:

$Height = 12cm$

$Radius = 6cm$

Step-by-step explanation:

Given

Represent volume with v, height with h and radius with r

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Required

Determine the values of h and r that uses the least amount of material

Volume is calculated as:

$V = \pi r^2h\\$

Substitute 432π for V

$432\pi = \pi r^2h$

Divide through by π

$432 = r^2h$

Make h the subject:

$h = \frac{432}{r^2}$

Surface Area (A) of a cylinder is calculated as thus:

$A=2\pi rh+2\pi r^2$

Substitute $\frac{432}{r^2}$ for h in $A=2\pi rh+2\pi r^2$

$A=2\pi r(\frac{432}{r^2})+2\pi r^2$

$A=2\pi (\frac{432}{r})+2\pi r^2$

Factorize:

$A=2\pi (\frac{432}{r} + r^2)$

To minimize, we have to differentiate both sides and set $A' = 0$

$A'=2\pi (-\frac{432}{r^2} + 2r)$

Set $A' = 0$

$0=2\pi (-\frac{432}{r^2} + 2r)$

Divide through by $2\pi$

$0= -\frac{432}{r^2} + 2r$

$\frac{432}{r^2} = 2r$

Cross Multiply

$2r * r^2 = 432$

$2r^3 = 432$

Divide through by 2

$r^3 = 216$

Take cube roots of both sides

$r = \sqrt[3]{216}$

$r = 6$

Recall that:

$h = \frac{432}{r^2}$

$h = \frac{432}{6^2}$

$h = \frac{432}{36}$

$h = 12$

Hence, the dimension that requires the least amount of material is when

$Height = 12cm$

$Radius = 6cm$

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