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3 years ago

8

What is the product? (5r + 2)(3r − 4)

1 answer:

Anarel [89]3 years ago

6 0

Distribute the first set of parenthesis

<span><span>15<span>r<span><span>2</span><span></span></span></span>−20r+6r−8</span><span>
Combine like terms

</span><span><span>15<span>r<span><span>2</span><span></span></span></span>+(−20r+6r)−8</span><span>
Simplify your answer

</span><span>15<span>r<span><span>2</span><span></span></span></span>−14r−8

*Answer is </span>15r^2−14r−8*<span><span>
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3 years ago

Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3

chubhunter [2.5K]

Answer:

The simplified form is $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

$\dfrac{2a^2-4a+2}{3a^2-3}$

Let us simplify the numerator and denominator separately.

The numerator is given as $2a^2-4a+2$

2 is a common factor in all the three terms. So, we factor it out. This gives,

$=2(a^2-2a+1)$

Now, $a^2-2a+1=(a-1)(a-1)$

Therefore, the numerator becomes $2(a-1)(a-1)$

The denominator is given as: $3a^2-3$

Factoring out 3, we get

$3(a^2-1)$

Now, $a^2-1$ is of the form $a^2-b^2=(a-b)(a+b)$

So, $a^2-1=(a-1)(a+1)$

Therefore, the denominator becomes $3(a-1)(a+1)$

Now, the given expression is simplified to:

$\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}$

There is $(x-1)$ in the numerator and denominator. We can cancel them only if $x\ne1$ as for $x=1$ , the given expression is undefined.

Now, cancelling the like terms considering $x\ne1$ , we get:

$\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}$

Therefore, the simplified form is $\dfrac{2(x-1)}{3(x+1)}$

The simplification is true only if $x\ne1$ . So, $x =1$ is the excluded value for the given expression.

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3 years ago

Solve. 3(x+1)-2x=-6

RideAnS [48]

D.) x = - 9

C.) x = 0

Hope that helps, Good luck! (:

4 0

3 years ago