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3 years ago

8

The table shows the steps for solving the given inequality for x. −2(3−2x)

1 answer:

Delicious77 [7]3 years ago

4 0

Answer:

3x = 4-x "hope this helps"

Step-by-step explanation:

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What is the value of n in the equation 1/2(n – 4) – 3 = 3 – (2n + 3)?

Travka [436]

Distribute 1/2 to <span>(n – 4):

</span> $\frac{1}{2} \times n = \frac{1}{2}n$
$\frac{1}{2} \times -4 = -2$
<span>
Subtract (2n + 3):

</span> $- (2n + 3) = -2n -3$
<span>
Your equation should now look like this:

</span> $\frac{1}{2}n - 2 - 3 = 3 - 2n - 3$
<span>
Combine like terms on both sides:

</span> $-2 - 3 = -5$
$3 - 3 = 0$
$\frac{1}{2}n - 5 = -2n$
<span>
Subtract 1/2n from both sides:

</span> $-5 = -\frac{5}{2}n$
<span>
Divide both sides by -5/2 to get n by itself:

</span> $n = 2$
<span>
The value of n is 2.</span>

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3 years ago

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Which of the following is true of the location of the terminal side of an angle, Theta, whose sine value is One-half?

FinnZ [79.3K]

Answer:

Theta has a reference angle of 30° and is in Quadrant I or II

Step-by-step explanation:

Sin(theta) = ½

Basic angle: 30

Angles:

30,

180-30 = 150

Because sin is positive in quadrants 1 and 2

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3 years ago

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81x^4 write as a square

Dahasolnce [82]

Answer:

a square what

Step-by-step explanation:

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3 years ago

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Match the inequality to the graph of its solution.

Tema [17]

The first graph is c. c - 7 < 3.
The second graph is a. c + 7 ≤ 3.
The third graph is b. c - 3 > 1.

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3 years ago

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Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

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3 years ago

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