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sergejj [24]
3 years ago
8

MARKNING BRAINLEST ! :) * The table shows a linear function, find the values of A , B , & C. Show your work.

Mathematics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

A = 11, B = 25/3, C = -1

Step-by-step explanation:

A linear function has the form:

f(x) = ax + b

(1)

f(3) = 8a + b = 8

f(5) = 5a + b = 9

Using the last 2 equations we can solve for 'a' and 'b'.

8a + b = 8 | 5a + b = 9

We multiply the second one by -1:

8a + b = 8 | -5a -b = -9

And then we add them together:

3a = -1

a = \frac{-1}{3}

We then solve for 'b':

5a + b = 9

5(-1/3) + b = 9

b = 9 + 5/3

b = 32/3 = 11\frac{1}{3}.

We then use this to find A, B, C.

f(x) = (-1/3)x + 32/3

(2)

f(A) = -A/3 + 32/3 = 7

\frac{-A}{3} + \frac{32}{3} = 7\\A - 32 = -21\\A = 11

(3)

f(7) = -7/3 + 32/3 = B

25/3 = B

(4)

f(C) = -C/3 + 32/3 = 11

\frac{-C}{3} + \frac{32}{3} = 11\\C - 32 = -33\\C = -1

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aleksandr82 [10.1K]
<span>the correct equation is
- 4j² + 3j - 28 =0

 ax² + bx + c,
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8 0
3 years ago
If f(x) =-2+8 and g(x) = square root of x+9 which statement is true
lidiya [134]

The statement that -6 is in the domain of f(g(x)) is true

<h3>Complete question</h3>

If f(x) = -2x + 8 and g(x) = \sqrt{x + 9, which statement is true?

  • -6 is in the domain of f(g(x))
  • -6 is not in the domain of f(g(x))

<h3>How to determine the true statement?</h3>

We have:

f(x) = -2x + 8

g(x) = \sqrt{x + 9

Start by calculating the function f(g(x)) using:

f(g(x)) = -2g(x) + 8

Substitute g(x) = \sqrt{x + 9

f(g(x)) = -2\sqrt{x + 9} + 8

Set the radicand to at least 0

x + 9 \ge 0

Subtract 9 from both sides

x \ge -9

This means that the domain of f(g(x)) are real numbers greater than or equal to -9. i.e. -9, -8, -7, -6, ...........

Hence, the statement that -6 is in the domain of f(g(x)) is true

Read more about domain at:

brainly.com/question/24539784

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3 0
2 years ago
If cot theta= 4/3, find csc theta
Marrrta [24]

Answer: Option d.

Step-by-step explanation:

The trigonometric identity needed is:

csc^2\theta=cot^2\theta+1

Knowing that cot\theta=\frac{4}{3}:

Substitute it into csc^2\theta=cot^2\theta+1:

csc^2\theta=(\frac{4}{3})^2+1

Simplify the expression:

csc^2\theta=(\frac{4}{3})^2+1\\\\csc^2\theta=\frac{16}{9}+1\\\\csc^2\theta=\frac{25}{9}

Solve for csc\theta. Apply square root at both sides of the expression:

\sqrt{csc^2\theta}=\±\sqrt{\frac{25}{9}}

csc\theta=\frac{5}{3}

8 0
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7 0
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Hope this helps have a nice day!
8 0
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