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4 years ago

How do you do pythagorean theorem when you need to fine the A

1 answer:

5 0

Given that Pythagorean Theorum is
A^2 + B^2 = C^2, you need to do the opposite.
To find for A^2, you need to:
Subtract B^2 from C^2. This creates the equation of:
C^2 - B^2 = A^2. This works because A^2 + B^2 is C^2, so the opposite must work.
Now if you're looking for A and not A^2, you need to subtract B^2 from C^2 and then square root the difference.
I hope this helps!

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If his conclusion is correct, which substance could he have been observing and why?

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3 years ago

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Find the first four terms of the sequence given by the following. an=43-6(n-1), n=1, 2, 3.

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Step-by-step explanation:

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3 years ago

a school cafeteria makes 4 diffrent salads during the week but serves only 2 salads each day on a roating basis. salads: chicken

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4 years ago

Find the z score for which the area to the right is 0.27.

Flura [38]

To find the Z score for a number to the right,

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1-0.27 = 0.73

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3 years ago

A storage box with a square base must have a volume of 90 cubic centimeters. the top and bottom cost $0.60 per square centimeter

Paul [167]

Answer-

For side length of 3.56 cm and height of 7.10 cm the cost will be minimum.

Solution-

Let us assume that,

x represents the length of the sides of the square base,

y represent the height.

Given the volume of the box is 90 cm³, so

$\Rightarrow V=90\\\\\Rightarrow x^2\times y=90\\\\\Rightarrow y=\dfrac{90}{x^2}$

As the top and bottom cost $0.60 per cm² and the sides cost $0.30 per cm². Total cost C will be,

$C=\text{cost for top and bottom}+\text{cost for rest 4 sides}\\\\=(2x^2\times 0.6)+(4xy\times 0.3)\\\\=(2x^2\times 0.6)+(4x\times \frac{90}{x^2}\times 0.3)\\\\=1.2x^2+ \dfrac{108}{x}$

Then,

$C'=\dfrac{d}{dx}(1.2x^2+ \dfrac{108}{x})=2.4x-\dfrac{108}\\\\C''=\dfrac{d^2}{dx^2}(1.2x^2+ \dfrac{108}{x})=2.4+\dfrac{216}{x^3}$

As C'' has all positive terms so, for every positive value of x (as length can't be negative), C'' is positive.

So, for minima C' = 0

$\Rightarrow 2.4x-\dfrac{108}{x^2}=0\\\\\Rightarrow 2.4x=\dfrac{108}{x^2}\\\\\Rightarrow x^3=\dfrac{108}{2.4}=45\\\\\Rightarrow x=3.56$

Then,

$y=\dfrac{90}{x^2}$

$y=\dfrac{90}{3.56^2}$

$y=7.10$

Therefore, for side length of 3.56 cm and height of 7.10 cm the cost will be minimum.

4 0

4 years ago

Read 2 more answers