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3 years ago

Write a statement that indicates that the triangles in each pair are congruent.

Mathematics

1 answer:

monitta3 years ago

4 0

A B because it’s included on the low end

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What is x when 2(x+3)=x-4

denpristay [2]

Answer:

$x=-10$

Step-by-step explanation:

First, use distribution to simplify

$2(x+3)=x-4\\2x+6=x-4$

Then, simplify further by combining like terms

$2x+6=x-4\\2x+10=x\\10=-1x$

Solve

$10=-1x\\x=-10\\$

8 0

3 years ago

Molly made 3600 \text{ mL}3600 mL3600, space, m, L of tea for a party, and she served the tea divided equally in 121212 cups.

kvasek [131]

Answer:

0.3

Step-by-step explanation:

3 0

3 years ago

If-2a > 6, then<br> -a>3<br> -a<3<br> -a<-3<br> -a>-3

chubhunter [2.5K]

Answer:

a < - 3

Step-by-step explanation:

Given

- 2a > 6

Divide both sides by - 2, reversing the symbol as a result of dividing by a negative quantity, thus

a < - 3

5 0

3 years ago

Which of the following is the best buy?

-BARSIC- [3]

The $1.25 one for 9 Oz box of rice would be the best buy.

<h3>How to find the Unit Price?</h3>

To get the best buy among the given items and prices, we need to find the unit price. Thus, the formula is;

Unit price = Price/Quantity

A) Unit Price = 0.90/6 = 0.15

B) Unit Price = 1.93/12 = 0.15083

C) Unit price = 2.42/16

Unit Price = 0.15125

D) Unit Price = 1.25/9 = 0.138

The $1.25 one for 9 Oz box of rice would be the best buy.

Read more about Unit price at; brainly.com/question/2804909

#SPJ1

3 0

2 years ago

Plz..... Solve this questions for me...

faust18 [17]

$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

$8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10$

3 0

3 years ago