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lbvjy [14]
3 years ago
7

Write an equation for a perpendicular line to the equation y = -1/2x + 7 and that goes through the point (2, 5).

Mathematics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

y = 2x + 1

Step-by-step explanation:

A line y = ax + b that is perpendicular to the other line y = (-1/2)x + 7

=> The product of 2 slopes of these 2 lines must be -1

=> a x (-1/2) = -1

=> a = 2

This line passes (2, 5) => 5 = 2*2 + b => 5 = 4 + b => b = 1

=> The equation of the line: y = 2x + 1

Hope this helps!

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Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. (2 points)
n200080 [17]

Given:

f(x)=\dfrac{8}{x}

g(x)=\dfrac{8}{x}

To find:

Whether f(x) and g(x) are inverse of each other by using that f(g(x)) = x and g(f(x)) = x.

Solution:

We know that, two function are inverse of each other if:

f(g(x))=x and g(f(x))

We have,

f(x)=\dfrac{8}{x}

g(x)=\dfrac{8}{x}

Now,

f(g(x))=f(\dfrac{8}{x})              [\because g(x)=\dfrac{8}{x}]

f(g(x))=\dfrac{8}{\dfrac{8}{x}}            [\because f(x)=\dfrac{8}{x}]

f(g(x))=8\times \dfrac{x}{8}

f(g(x))=x

Similarly,

g(f(x))=f(\dfrac{8}{x})              [\because f(x)=\dfrac{8}{x}]

g(f(x))=\dfrac{8}{\dfrac{8}{x}}            [\because g(x)=\dfrac{8}{x}]

g(f(x))=8\times \dfrac{x}{8}

g(f(x))=x

Since, f(g(x))=x and g(f(x)), therefore, f(x) and g(x) are inverse of each other.

6 0
3 years ago
Solve <br><br> 9(a + 2b) + c <br><br> a = 3 <br> b = 2<br> c = 1
swat32

Answer:

64

Step-by-step explanation:

Step 1: Define

9(a + 2b) + c

a = 3

b = 2

c = 1

Step 2: Substitute and Evaluate

9(3 + 2(2)) + 1

9(3 + 4) + 1

9(7) + 1

63 + 1

64

4 0
3 years ago
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zepelin [54]

Answer:

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Step-by-step explanation:

5 0
3 years ago
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olga nikolaevna [1]

Answer:

3^{-2} = \dfrac{1}{9}

Step-by-step explanation:

Definition of negative exponent:

a^{-n} = \dfrac{1}{a^n}

In our case, a = 3 and -n = -2.

3^{-2} = \dfrac{1}{3^2}

Now we just evaluate the power in the denominator.

3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}

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Answer:

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Step-by-step explanation:

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