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mario62 [17]
2 years ago
5

Which ordered pair makes both inequalities true?

Mathematics
1 answer:
Tatiana [17]2 years ago
3 0

The ordered pair which makes both inequalities true is: D. (3, 0).

<h3>How to determine ordered pair?</h3>

In Mathematics, an inequality can be used to show the relationship between two (2) or more integers and variables in an equation.

In order to determine ordered pair which makes both inequalities true, we would substitute the points into the inequalities as follows:

At (0, 0), we have:

y > -2x + 3  

0 > -2(0) + 3

0 > 3 (false).

y < x – 2

0 < 0 - 2

0 < -2 (false)

At (0, -1), we have:

y > -2x + 3  

-1 > -2(0) + 3

-1 > 3 (false).

y < x – 2

-1 < 0 - 2

-1 < -2 (false)

At (1, 1), we have:

y > -2x + 3  

1 > -2(1) + 3

1 > -1 (true).

y < x – 2

1 < 1 - 2

1 < -1 (false)

At (3, 0), we have:

y > -2x + 3  

0 > -2(3) + 3

0 > -3 (true).

y < x – 2

0 < 3 - 2

0 < 1 (true).

Read more on inequalities here: brainly.com/question/24372553

#SPJ1

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7 0
3 years ago
If f(x)=x^3-12x^2+35x-24f(x)=x 3 −12x 2 +35x−24 and f(8)=0f(8)=0, then find all of the zeros of f(x)f(x) algebraically.
Neporo4naja [7]

Answer:

The zeros of f(x) are: (x - 1), (x - 3) and (x - 8)

<em></em>

Step-by-step explanation:

Given

f(x)=x^3-12x^2+35x-24

f(8) = 0

Required

Find all zeros of the f(x)

If f(8) = 0 then:

x = 8

And x - 8 is a factor

Divide f(x) by x - 8

\frac{f(x)}{x - 8} = \frac{x^3-12x^2+35x-24}{x - 8}

Expand the numerator

\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 -8x^2 + 3x + 32x - 24}{x - 8}

Rewrite as:

\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 + 3x - 8x^2 +32x - 24}{x - 8}

Factorize

\frac{f(x)}{x - 8} = \frac{(x^2 - 4x + 3)(x - 8)}{x - 8}

Expand

\frac{f(x)}{x - 8} = \frac{(x^2 -x - 3x + 3)(x - 8)}{x - 8}

Factorize

\frac{f(x)}{x - 8} = \frac{(x - 1)(x - 3)(x - 8)}{x - 8}

\frac{f(x)}{x - 8} = (x - 1)(x - 3)

Multiply both sides by x - 8

f(x) = (x - 1)(x - 3)(x - 8)

<em>Hence, the zeros of f(x) are: (x - 1), (x - 3) and (x - 8)</em>

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