What is the solution to the inequality?

2 answers:

8 0

The steps to solving an inequality are: add or subtract from each side - multiply or divide both sides - simplify.

5x + 8 > -12
5x > -12 - 8
5x > -20
x > -20/5
x > -4

The answer is: x > -4

Liono4ka [1.6K]3 years ago

8 0

5x + 8 > -12 5x > -12-8 . 5x> -20 . 5/5x> -20/5 . x > -4 C

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What are the coordinates of Point X?<br> A. (3, –4)<br> B. (–3, –4)<br> C. (4, –3)<br> D. (4, 3)

Greeley [361]

Answer:

3, -4

Step-by-step explanation:

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8 0

3 years ago

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Mario bought $5 worth of stamps at the post office. He bought ten more 6 cents stamps than 10 cents stamps. The number of 8 cent

vlabodo [156]

S=number of 6 cent stamps
t=number of ten cen stamps
e=number of 8 cent stamps
w=number of 20 cents stamps=2

cost=500 cents
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also
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8 cent stamps is 3 times 10 cent stamps
e=3t

sub
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3(10+t)+5t+4(3t)+230
30+3t+5t+12t=230
30+20t=230
minus 30
20t=200
divid 20
t=10

sub back
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s=10+10
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e=3(10)
e=30

20 six cent
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10 ten cent
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6 0

3 years ago

Read 2 more answers

Use the Factor Theorem and synthetic division to show x + 5 is a factor of f(x) = 2x3 + 7x2 − 14x + 5

olga2289 [7]

Answer:

Factor theorem: $f(-5) = 0$ .

Synthetic division: $f(x) = (x + 5)\, (2\, x^2 -3\, x + 1)$ .

Step-by-step explanation:

<h3>Factor Theorem</h3>

By the factor theorem, a monomial of the form $(x - a)$ (where $a$ is a constant) is a factor of polynomial $f(x)$ if and only if $f(a) = 0$ .

In this question, the monomial is $(x + 5)$ , which is equivalently $(x - (-5))$ . $a = -5$ .

$\begin{aligned}& f(-5) \\ &= 2 \times (-5)^{3} + 7 \times (-5)^{2}- 14\times (-5) + 5\\ &= -250 + 175 + 70 + 5 \\ &= 0\end{aligned}$ .

Hence, by the factor theorem, $(x + 5)$ , which is equivalent to $(x - (-5))$ , is a factor of $f(x)$ .

<h3>Synthetic Division</h3>

$\begin{aligned}& f(x) \\ &= 2\, x^{3} + 7\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} - 3\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} + \underbrace{(x + 5)\, (-3\, x)}_{-3\, x^{2} - 15\, x} + (x + 5) \\ &= (x + 5)\, (2\, x^{2} - 3\, x + 1)\end{aligned}$ .