Write and solve an inequality for each problem. Jorge’s soccer team is having its annual fundraiser. The team hopes

Mathematics

1 answer:

adelina 88 [10]3 years ago

5 0

First of all, they need to triple their profits. So 87 x 3 = 241.

When you read the question, it says at least. In inequality terms, that means greater than equal to. You’d do this:

x is the amount of money they raise this year.

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Directions : Factor each of the following Differences of two squares and write your answer together with solution

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$\huge \boxed{\mathfrak{Question} \downarrow}$

Factor each of the following differences of two squares and write your answer together with solution.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf \: x ^ { 2 } - 36$

Rewrite $\sf\:x^{2}-36 as x^{2}-6^{2}$ . The difference of squares can be factored using the rule: $\sf\: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\boxed{ \boxed{ \bf\left(x-6\right)\left(x+6\right) }}$

__________________

$\sf \: 49 - x ^ { 2 }$

Rewrite 49-x² as 7²-x². The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\sf \: \left(7-x\right)\left(7+x\right)$

Reorder the terms.

$\boxed{ \boxed{ \bf\left(-x+7\right)\left(x+7\right) }}$

__________________

$\sf \: 81 - c ^ { 2 }$

Rewrite 81-c²as 9²-c². The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\sf\left(9-c\right)\left(9+c\right)$

Reorder the terms.

$\boxed{ \boxed{ \bf\left(-c+9\right)\left(c+9\right) }}$

__________________

$\sf \: m ^ { 2 } n ^ { 2 } - 1$

Rewrite m²n² - 1 as $\sf\left(mn\right)^{2}-1^{2}$ . The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .