Choose the two equations you would use to solve the absolute value equation below. Then solve the two equations.|x + 5| = 55

Answers

Answer 1

Answer: $|x+5|=55\n\nx+5=55\ \vee\ x+5=-55\n\nx=55-5\ \vee\ x=-55-5\n\nx=50\ \vee\ x=-60$

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What is the product of (x + 3)(2x – 1)

Answers

Answer:

The product of $(x + 3)(2x -1)=2x^2+5x-3$

Step-by-step explanation:

Given : Two expressions (x + 3) and (2x – 1)

We have to find the product of given expression that is $(x + 3)(2x -1)$

Consider the given expression $(x + 3)(2x -1)$

Multiply each term of first bracket with each term of second bracket

$\left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

we have,

$=x\cdot \:2x+x\left(-1\right)+3\cdot \:2x+3\left(-1\right)$

Simplify, $-(-a)=+a$

we have $=2xx-1\cdot \:x+3\cdot \:2x-3\cdot \:1$

Simplify, we get,

$=2x^2-x+6x-3$

Adding like terms, we have,

$=2x^2+5x-3$

Thus, the product of $(x + 3)(2x -1)=2x^2+5x-3$

Umm
(X+3)(2x-1)
=2x square -x + 6x - 3
=2x square +5x -3

Which doesnt belong and why

Answers

Answer:

Step-by-step explanation:

They all have and addition and subtraction pattern in each cube, thank me later - PrObLeM OcCuReD

Roger wraps presents at a local gift shop. If it takes 1.4 meters of ribbon to wrap a present, how many can he wrap if he has 40 meters of ribbon?

Answers

40/1.4= 28.57 Reasonably, he can wrap 28 gifts before running out of ribbon.

There are 15 mill teachers in the school the 35 female teachers in the school use a ratio

Answers

The ratio of male: female teachers
15:35
On simplifying;
3:7

1. Find the Least Common Multiple of these two monomials:
See picture

Answers

Answer:

The last choice is correct

$LCM=120a^4b^7c^5$

Step-by-step explanation:

Least Common Multiple (LCM)

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

$12a^4b^2c^5$

$40a^3b^7c^1$

The prime factors of the first monomial are:

$2^2,3,a^4,b^2,c^5$

The prime factors of the second monomial are:

$2^3,5,a^3b^7c^1$

LCM = Multiply $2^3*3*5*a^4*b^7*c^5$

These are all the factors the greatest number of times they occur.

Operating:

$LCM=8*15*a^4*b^7*c^5$

$\boxed{LCM=120a^4b^7c^5}$

The last choice is correct

PLZZZ ANSWERR 20 points Select correct answerIn the figure, angle A measured 41* and angle D measures 32*. What is the he measurements of angle E

Answers

It’s 10 points not 20
Explanation

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