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-BARSIC- [3]
3 years ago
12

Let f and g be functions. (a) The function (f + g)(x) is defined for all values of x that are in the domain(s) of ____________.

(b) The function (fg)(x) is defined for all values of x that are in the domain(s) of __________. (c) The function (f/g)(x) is defined for all values of x that are in the domain(s) of__________
Mathematics
1 answer:
krok68 [10]3 years ago
5 0

Answer:

a) for all values of x that are in the domains of f and g.

b) for all values of x that are in the domains of f and g.

c) for all values of x that are in the domains of f and g with g(x)≠0

Step-by-step explanation:

a) By definition (f+g)(x)=f(x)+g(x). Then x must be in the domain of f and g.

b) By definition (fg)(x)=f(x)g(x).  Then x must be in the domain of f and g.

c) By definition (f/g)(x)=f(x)/g(x).  Then x must be in the domain of f and g and g(x) must be different of 0.

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8x over 3 - x over 2 equals -13
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Equation:
x/3 = x/8 + 2/3
---
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========

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3 years ago
Solve each inequality, and then drag the correct solution graph to the inequality.
Nesterboy [21]

The correct solution graph to the inequalities are

4(9x-18)>3(8x+12)  →  C

-\frac{1}{3}(12x+6) \geq -2x +14  → A

1.6(x+8)\geq 38.4  →  B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

4(9x-18)>3(8x+12)

First, clear the brackets,

36x-72>24x+36

Then, collect like terms

36x-24x>36+72\\12x >108

Now divide both sides by 12

\frac{12x}{12} > \frac{108}{12}

∴ x > 9

For the second inequality

-\frac{1}{3}(12x+6) \geq -2x +14

First, clear the fraction by multiplying both sides by 3

3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)

-1(12x+6) \geq -6x +42

Now, open the bracket

-12x-6 \geq -6x +42

Collect like terms

-6 -42\geq -6x +12x

-48\geq 6x

Divide both sides by 6

\frac{-48}{6} \geq \frac{6x}{6}

-8\geq x

∴ x\leq  -8

For the third inequality,

1.6(x+8)\geq 38.4

First, clear the brackets

1.6x + 12.8\geq 38.4

Collect likes terms

1.6x \geq 38.4-12.8

1.6x \geq 25.6

Divide both sides by 1.6

\frac{1.6x}{1.6}\geq  \frac{25.6}{1.6}

∴ x \geq  16

Let the graphs be A, B and C from left to right

The first graph (A) shows x\leq  -8 and this matches the 2nd inequality

The second graph (B) shows x \geq  16 and this matches the 3rd inequality

The third graph (C) shows x > 9 and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

4(9x-18)>3(8x+12)  →  C

-\frac{1}{3}(12x+6) \geq -2x +14  → A

1.6(x+8)\geq 38.4  →  B

Learn more here: brainly.com/question/17448505

8 0
2 years ago
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