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3 years ago

What the answer to this

Mathematics

1 answer:

neonofarm [45]3 years ago

6 0

Answer:

Step-by-step explanation:

(I solved in the picture, firstly look at there)

after completing the square and factoring :

${x}^{2} + 4x + 4 = 8 + 4$

${x}^{2} + 4x + 4 = 12$

${(x - 2)}^{2} = 12$

Hope this helps ^-^

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Simplify with steps.

gogolik [260]

1. (x−1)(5x+8)−2x+5

=(x)(5x)+(x)(8)+(−1)(5x)+(−1)(8)+−2x+5

=5x2+8x+−5x+−8+−2x+5

Combine Like Terms:

=5x2+8x+−5x+−8+−2x+5

=(5x2)+(8x+−5x+−2x)+(−8+5)

=5x2 + x + −3

2. (−2+5)(3x+9)

=9x+27

3. 2x+5−8x+12

=2x+5+−8x+12

Combine Like Terms:

=2x+5+−8x+12

=(2x+−8x)+(5+12)

=−6x+17

4. (−x+3)(−5x−2)

=(−x+3)(−5x+−2)

=(−x)(−5x)+(−x)(−2)+(3)(−5x)+(3)(−2)

=5x2+2x−15x−6

=5x2 − 13x − 6

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3 years ago

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4 years ago

Undefined undefined 0 0 2

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3 years ago

Look at the graph below Which part of the graph best represents the solution set to the system of inequalities y ≥ x + 1 and y +

kotykmax [81]

ANSWER

Part A

EXPLANATION

The given inequalities are,

$y \geqslant x + 1$

and

$y + x \geqslant - 1$

To see which part of the graph best represent the solution set, choose a point from each part and substitute in to the inequalities.

If a point from a given part satisfies the inequalities simultaneously, then that part best represents the solution set.

Part A.

We choose
$(0,2)$
We plug in to the inequalities.

$2 \geqslant 0 + 1$

$\Rightarrow \: 2 \geqslant 1$
The above inequality is true.

We plug in to the second inequality.

$2 + 0 \geqslant - 1$

.
$\Rightarrow \: 2 \geqslant - 1$

This statement is also true.

Part B.

If we plug in
$(-2,0)$
in to the first statement, we get,

$0 \geqslant - 2 + 1$

This implies that,

$0 \geqslant - 1$

This is true.

If substitute in to the second, we get,

$0 + - 2\geqslant - 1$

$\Rightarrow \: - 2 \geqslant - 1$
This is false.

Part C

We plug
$(0,-2)$
in to the first inequality

$- 2 \geqslant 0 + 1$
This means that,

$- 2 \geqslant 1$
This is false.

We plug in to the second inequality,

$- 2 + 0 \geqslant -1$

$- 2 \geqslant -1$

False.

Part D also has the point

$(2,0)$

We put this point in to the first inequality to get,

$0 \geqslant 2 + 1$
$0 \geqslant 3$
This is false.

Then in to the second inequality.

$0 + 2 \geqslant 1$
$2 \geqslant -1$

This final statement is true.

Since the point from Part A satisfies both inequalities simultaneously, it represents the solution set.

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4 years ago

Read 2 more answers

What is m<2 + m<3? Explain.

just olya [345]

Answer:

$\large\boxed{m$

Step-by-step explanation:

$m$

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3 years ago