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Ivenika [448]
2 years ago
14

easy algebra question below first correct answer gets brainliest, if you put a link i will report and block

Mathematics
1 answer:
boyakko [2]2 years ago
3 0

Answer:

6x²

Step-by-step explanation:

3x×2x=6...... then an x²

Therefore answer is 6x²

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Look at the table. Make a conjecture about the sum of the first 10 positive even numbers.
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The answer is 10•11, if you look at the pattern, you don't have to solve the addition.

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Which is the best deal<br>12oz. for $.99<br>64oz for $2.99<br>128 oz. $4.99​
mojhsa [17]

Answer:

best deal is 64oz for $2.99

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What is the domain of g(x) = sqrt x +1
Shkiper50 [21]
There can be no negative in a sqrt 
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x+1\geq 0
x \geq -1
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7 0
3 years ago
Using the distributive property to find the product (y – 4)(y2 + 4y + 16) results in a polynomial of the form y3 + 4y2 + ay – 4y
Crank

Answer:

The answer to your question is a = 16

Step-by-step explanation:

Polynomial

                  (y - 4) (y² + 4y + 16)

Process

1.- Multiply y by each term of the polynomial

                y(y² + 4y + 16) = y³ + 4y² + 16y

2.- Multiply -4 by each term of the polynomial

                -4(y² + 4y + 16) = -4y² - 16y - 64

3.- Write both results

                y³ + 4y² + 16y - 4y² - 16y - 64

In bold we notice that a = 16

8 0
3 years ago
What is the solution set for 2x+5y&gt;-1 and 4x-3&lt;-3?
bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}

\displaystyle\ x > \frac{-1 - 5y}{2}

So, the solution for x in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 - 5y}{2}

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}

\displaystyle\ x > \frac{-1 -1x}{5}

So, the solution for y in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 -1x}{5}

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

\displaystyle\frac{4x}{4}

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

•

- <em>Marlon Nunez</em>

7 0
3 years ago
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