Find the perimeter of the triangle or rectangle. The sides of each triangle are equal in length. Could you help with 16 and 18?

Mathematics

1 answer:

Whitepunk [10]3 years ago

8 0

16. 9

Since each side are congruent, you set the two sides to each other

7x-12=3x

Subtract 7x from both sides

-12=-4x

Divide -4 from both sides

3=x

3+3+3=9

18. 28

2x-3=3x-10

Add 10 on both sides

2x+7=3x

Subtract 2x on both sides

7=x

7+7+7+7=28

You might be interested in

Simplify. 18x + 3xy + 5y −2(6xy − 2y + 4) A) 18x − 9xy + 9y − 8 B) 18x + 9xy + 9y − 8 C) 18x + 9xy − 9y + 8 D) 18x − 9xy + y − 8

Karolina [17]

18x+3xy+5y-2(6xy-2y+4)

18x+3xy+5y-12xy+4y-8

18x-9xy+9y-8

The answer is D

8 0

3 years ago

Read 2 more answers

What is the estimated difference? Round each number to the nearest hundred thousand before finding the difference? 3,661,528 – 1

solmaris [256]

3,700,000-1,700,000=2,000,000

6 0

3 years ago

Joyce is comparing two different internet providers to see which is a better deal. The cost of service from Provider S can be re

maxonik [38]

This means an initial fee of $45 have to be paid to the provider and a charge of $35 per month.

<h3>Linear equation</h3>

A linear equation is in the form:

y = mx + b

Where y, x are variables, m is the rate of change and b is the initial value of y.

Let S represent the total cost of service for x months. Given:

y = 35x + 45

This means an initial fee of $45 have to be paid to the provider and a charge of $35 per month.

Find out more on Linear equation at: brainly.com/question/13763238

5 0

2 years ago

Add and Subtract Rational Expressions with a Common Denominator

Sladkaya [172]

Answer:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}$

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}$