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Black_prince [1.1K]
3 years ago
13

Question 1 Solve the absolute value equation |x| − 2 = 5 to find the distance Mary can walk forward, x, in 5 seconds. Group of a

nswer choices x = 7 m and x = −7 m x = −7 m and x = 3 m x = −3 m and x = 7 m x = 3 m and x = −3 m
Mathematics
1 answer:
Darya [45]3 years ago
3 0

Given:

The absolute value equation is

|x|-2=5

To find:

The distance Mary can walk forward, i.e., x.

Solution:

We have,

|x|-2=5

Add 2 on both sides.

|x|-2+2=5+2

|x|=7

x=\pm 7

x=-7 and x=7

Here, negative sign represents the distance Mary can walk reverse direction.

Therefore, the correct option is A.

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Ahhhh I see now yes I do
user100 [1]

Answer: yes

Step-by-step explanation: so smart I think

4 0
3 years ago
Select the correct answer. Which set of coordinates satisfies the equations x − 2y = -1 and 2x + 3y = 12? A. (1, 2) B. (3, 1) C.
alekssr [168]

Answer:

The coordinates that satisfy the equations will be: (3, 2)

Hence, <em>option C</em> is correct.

Step-by-step explanation:

Given the equations

x\:-2y\:=-1;\:2x\:+\:3y=12

solving the equations to find the solution values (coordinates)

\begin{bmatrix}x-2y=-1\\ 2x+3y=12\end{bmatrix}

\mathrm{Multiply\:}x-2y=-1\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:2x-4y=-2

\begin{bmatrix}2x-4y=-2\\ 2x+3y=12\end{bmatrix}

2x+3y=12

-

\underline{2x-4y=-2}

7y=14

so

\begin{bmatrix}2x-4y=-2\\ 7y=14\end{bmatrix}

solve for y

7y=14

y=2

\mathrm{For\:}2x-4y=-2\mathrm{\:plug\:in\:}y=2

2x-4\cdot \:2=-2

2x=6

\mathrm{Divide\:both\:sides\:by\:}2

\frac{2x}{2}=\frac{6}{2}

x=3

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=3,\:y=2

Therefore, the coordinates that satisfy the equations will be: (3, 2)

Hence, <em>option C</em> is correct.

7 0
3 years ago
Read 2 more answers
A.
anygoal [31]
The Correct answers are
A. 1.3
E. -1
F. -4
8 0
3 years ago
I need help with my math
Zepler [3.9K]
OK, what is it? I'd love to help.

3 0
4 years ago
Please help anyone know how to do this.
stira [4]

A right angle triangle will follow Pythagorean theory where a^2(hypotenuse)= b^2+ c^2

Therefore 17^2=11^2+ c^2

289=121 + c^2

168 = c^2

c= 13 rounded to the nearest tenth

7 0
3 years ago
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