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3 years ago

Evaluate the expression (5x^2)^3/(10y)^4 for x=10 and y=5

2 answers:

8 0

Pay attention to the procedure:
<span>(5(10)^2)^3/(10(5))^4. What we need is plug in the corresponding x and y values Then you have to do x squared before you multiply it by 5 like this: (5(100))^3/(10(5))^4 Now solve for what is in the top parenthesis. (500)^3/(10(5))^4 500^3=125000000 Now work on the denominator. Do what is inside the parenthesis first 10*5=50 Now (50)^4=6250000 Now divide the numerator and the denominator. 125000000/6250000 Which equals 20, Please check if I'm wrong but I think this is what you need </span>

zhenek [66]3 years ago

5 0

Answer: The answer is option D i.e. 20

Step-by-step explanation:

We have the expression

((5x^2) ^3)/(10y)^4

and values of x=10, y=5

Putting the values of x and y in the expression

((5×10^2)^3)/ (10×5)^4

= (5×100)^3/(50^4)

= (500^3)/6250000

125000000/6250000

= 20

So the expression (5x^2)^3/(10y)^4 is evaluated to 20

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3 years ago

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Read 2 more answers

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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3 years ago