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Novosadov [1.4K]
3 years ago
12

Solve the following and graph the solutions:

Mathematics
1 answer:
Oksanka [162]3 years ago
4 0
<h2>Solving Compound Inequalities</h2><h3>Answer:</h3>

The third Choice

x > 1 \text{ or } x < -1

<em>Please</em><em> refer</em><em> to</em><em> the</em><em> attached</em><em> image</em><em> </em><em>for</em><em> the</em><em> </em><em>graph</em>

<h3>Step-by-step explanation:</h3>

Given:

2x +1 > 3 \text{ or } 2x -1 < -3

Let's solve for the 2x +1 > 3 part of the Compound Inequality first.

2x +1 > 3 \\ 2x > 3 -1 \\ 2x > 2 \\ x > \frac{2}{2} \\ x > 1

Now, let's solve 2x -1 < -3:

2x -1 < -3 \\ 2x < -3 +1 \\ 2x < -2 \\ x < \frac{-2}{2} \\ x < -1

We can now rewrite our Compound Inequality as x > 1 \text{ or } x < -1

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I’m confused on this
dexar [7]

\huge\boxed{x=25}

The mistake is in Step 1. The solver should have also squared 9 when squaring the other side of the equation.

<h3>Correctly solving</h3>

\begin{aligned}\sqrt{3x+6}&=9\\(\sqrt{3x+6})^2&=9^2\\3x+6&=81\\3x+6-6&=81-6\\3x&=75\\\frac{3x}{3}&=\frac{75}{3}\\x&=25\end{aligned}

4 0
2 years ago
Read 2 more answers
Acute angle DOG with a measure of 45 degrees
Vlad1618 [11]
So make and angle with the measure of 45 degrees. And then label it DOG. It doesn't matter where you start labeling as long as O is your midpoint. :)
6 0
3 years ago
The radius of Circle A is 6 mm. The radius of Circle B is 4 mm greater than the radius of Circle A. The radius of Circle C is 2
labwork [276]

Answer:

Area of circle A =113.14 mm²

Area of circle b = 314.29 mm²

Area of circle C = 452.57 mm²

Area of circle A = 254.57 mm²

2.25 times

Step-by-step explanation:

Area of a circle = nr²

where n = 22/7

r = radius

Circle A's radius = 6mm

Circle B's radius = 6mm + 4mm = 10mm

Circle C's radius = 10mm + 2mm = 12mm

Circle D's radius = 12mm - 3mm = 9mm

Area of circle A = (22/7) x 6² = 113.14 mm²

Area of circle b = (22/7) x 10² = 314.29 mm²

Area of circle C = (22/7) x 12² = 452.57 mm²

Area of circle A = (22/7) x 9² = 254.57 mm²

Number of times the area of circle D is greater than that circle A = Area of circle D / Area of circle A

254.57 mm² / 113.14 mm² = 2.25 times

4 0
3 years ago
Help me solve those show work 3(2a-5)=12-7a, 27=3c-3(6-2c)and 6c-8-2c=-16
posledela
3(2a-5) = 12-7a
6a-15=12-7a
6a+7a=12+15
13a= 27

27=3c-3(6-2c)
27=3c-18+6c
27+18=3c+6c
45=9c
C=45/9
C=5

6c-8-2c=-16
6c-2c=-16+8
4c=-8
C=-8/4
C=-2
3 0
3 years ago
The intersection of a triangle and a line could be a segment<br> True or False
Arlecino [84]

false

                                                                                                                                                   vdvfvfddfvgggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg

6 0
3 years ago
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