All categories

3 years ago

Anyone help,easy questions ,10 points !!!

Mathematics

1 answer:

Usimov [2.4K]3 years ago

5 0

#21 the answer is does √ 9^2 + 2^2

a^2 + b^2 = c^2

9^2 + 2^2 = c^2

√ 9^2 + 2^2 = √ c^2

You might be interested in

Blake gives piano lessons in the evenings.

Aliun [14]

Answer:

The correct answer is B. {0, 20, 40, 60, 80, 100, 120}

Step-by-step explanation:

1. Let's check all the information given to answer this question correctly:

Rate of each piano lesson Blake gives = US$ 20

Blake can give between 0 and 6 lessons every evening

Earnings = Number of lessons * Rate of each lesson

2. What is the range of this relation?

Let's recall that the domain of a relation is the set of all values for which the relation is defined, and the range of the relation is the set of all values that the relation takes.

It means that the domain of the relation in this case is the number of lessons that Blake can give every evening:

{0, 1, 2, 3, 4, 5, 6}

And the range of this relation is the amount Blake earns in an evening depending on the number of lessons he gives:

{0, 20, 40, 60, 80, 100, 120}

The correct answer is B. {0, 20, 40, 60, 80, 100, 120}

3 0

3 years ago

Read 2 more answers

What are the terms of 12x^2

alexandr402 [8]

The terms of 12x^2=36=6

6 0

3 years ago

A series of trade discounts

Inessa05 [86]

they offer a lot of discounts prety much

8 0

3 years ago

Triangle ABC has vertices at A(−1 , 2), B(4, 2), and C(3, −1). Classify the triangle according to the side lengths.

Orlov [11]

Answer:

B) Isosceles

Step-by-step explanation:

The given triangle has vertices at A(-1,2), B(4,2) and C(3,-1).

We must first determine the length of the sides of the triangle, before we can classify it.

We apply the distance formula to find length of the sides.

$|AB|=\sqrt{(4--1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{5^2+(0)^2}$

$\Rightarrow |AB|=\sqrt{25}$

$\Rightarrow |AB|=5 units$ .

The length of side BC

$|BC|=\sqrt{(3-4)^2+(-1-2)^2}$

$\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}$

$\Rightarrow |BC|=\sqrt{1+9}$

$\Rightarrow |BC|=\sqrt{10}$

The length of side AC

$|AC|=\sqrt{(3--1)^2+(-1-2)^2}$

We simplify to obtain;

$|AC|=\sqrt{(3+1)^2+(-3)^2}$

$\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}$

$|AC|=\sqrt{16+9}$

$|AC|=\sqrt{25}$

$|AC|=5\:units$

Since $|AC|=5\:units=|AB|$ , the given triangle is an isosceles triangle.

The correct answer is

6 0

3 years ago

Read 2 more answers

2 Points Classify the following triangle. Check all that apply.

IgorLugansk [536]

The triangle is right

7 0

3 years ago

Read 2 more answers