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Andrei [34K]
3 years ago
15

Determine the value of variables a, b, and c that make each equation true.

Mathematics
1 answer:
dybincka [34]3 years ago
4 0

Corrected Question

Determine the values of a, b and c that make each equation true.

(x^a)^6=\dfrac{1}{x^{30}} \\\\(x^{-7})^{-4}=x^b\\\\(x^{-2})^c=x^{22}

Answer:

a=-5, b=28 and c=-11

Step-by-step explanation:

To solve for a,b and c, we apply the following laws of indices

\dfrac{1}{x^y}=x^{-y} \\\\(x^m)^n=x^{m X n}\\\\$If x^m=x^n,$ then m=n

Therefore

(x^a)^6=\dfrac{1}{x^{30}}\\x^{a*6}=x^{-30}\\6a=-30\\a=-5

To solve for b

(x^{-7})^{-4}=x^b\\x^{-7*-4}=x^b\\x^{28}=x^b\\b=28

To solve for c

(x^{-2})^c=x^{22}\\x^{-2*c}=x^{22}\\-2c=22\\c=-11

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Vanyuwa [196]
1. The correct statement is the first one, which is: <span>Add the two equations, solve for y, and then substitute –1 for y to find x.
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 2. Therefore, you have the following system of equations:
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6 0
3 years ago
Read 2 more answers
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8 0
3 years ago
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Maru [420]

Answer: f(x) = 3x^(2)-2x+1

f(x) = 3x^(2)-2x+1 = y-intercept (0, 1).

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As we see, the first equation (f(x) = 3x^(2)-2x+1) has a y-intercept of (0, 1), while the other equation (g(x) = -10x^(2)+50x) has a y-intercept of (0, 0). This means that the quadratic equation that has the greatest y-intercept is f(x) = 3x^(2)-2x+1, which is (0, 1).

Step-by-step explanation:

Hope this helps =)

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