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Charra [1.4K]
2 years ago
14

Rosa has $175 in her online music account. Each month she buys $15 worth of new songs. Which inequality could be solved

Mathematics
1 answer:
Virty [35]2 years ago
6 0

Answer:

it is C)15x - 175 < 10

Step-by-step explanation:

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Let f(x) =7x-13. Find f -1(x)
muminat
If you’re trying to find f(-1) it would be -20
5 0
3 years ago
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Given f (x ) = x^2 + 3x + 2 and g (x ) = x + 1, perform the indicated operations.
Nana76 [90]

Answer:

(i) (f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2

Step-by-step explanation:

The given functions are;

f(x) = x² + 3·x + 2

g(x) = x + 1

(i) (f - g)(x) = f(x) - g(x)

∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1

(f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3

(f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = f(x) × g(x)

∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2

(f·g)(x) = x³ + 4·x² + 5·x + 2

7 0
3 years ago
PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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3 years ago
What is the x-intercept of the line 5x+2y=10
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5x is the x-intercept.
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