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3 years ago

8

Amir wants to scale up a photo to create a poster for his room. The original picture is 6 inches in width and 4 inches in height

. What will be the height of the poster if its width is 30 inches?

1 answer:

kari74 [83]3 years ago

7 0

20 inches in height because 30 divided by 6 is 5 so you multiply 4 times 5

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If ADD = 9, BAD = 7, and CAD = 8, what is the value of ADA?

snow_lady [41]

give each letter a number: A=1, B=2, C=3, D=4

ADD = 1+4+4=9

BAD = 2+1+4 = 7

CAD = 3+1+4 = 8

So ADA = 1+4+1 = 6

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3 years ago

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20 points <br><br> What do you remember about types of angles?

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An acute angle is one that measures LESS than 90˚

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Have a great day <3

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3 years ago

Allison was left an inheritance from her great aunt on her 12 birthday. However, she , she not able to touch it until she is 21

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3 years ago

Simplify the radical expression

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3 years ago

A box with a square base and open top must have a volume of 157216 cm3. We wish to find the dimensions of the box that minimize

shepuryov [24]

Answer:

Base Length of 68cm
Height of 34 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume $=x^2h=157216$

$h=\dfrac{157216}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{157216}{x^2}\\A(x)=x^2+4x\left(\dfrac{157216}{x^2}\right)\\A(x)=\dfrac{x^3+628864}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+628864}{x}\\A'(x)=\dfrac{2x^3-628864}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68$

Step 4: Verify that x=68 is a minimum value

We use the second derivative test

$If\:A(x)=\dfrac{x^3+628864}{x}\\A''(x)=\dfrac{2x^3+1257728}{x^3}\\$When x=68\\A''(x)=6$

Since the second derivative is positive at x=68, then it is a minimum point.

Recall:

$h=\dfrac{157216}{x^2}\\h=\dfrac{157216}{68^2}=34$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 68cm
Height of 34 cm.

3 0

3 years ago