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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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3 years ago

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

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$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

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3 years ago