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Alexus [3.1K]
3 years ago
10

2 + 3(3x - 6) = 5(x - 3) + 15 I'm in Algebra 1 and I still can't seem to get this problem, I use Slader to find how to do this b

ut I still struggle, could someone please break this down, step by step
Mathematics
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

x = 4

Step-by-step explanation:

2 + 3(3x - 6) = 5(x - 3) + 15

Expand the parenthesis:

2 + 9x - 18 = 5x - 15 + 15

Simplify:

9x - 16 = 5x

Subtract 5x from both sides:

4x - 16 = 0

Add 16 to both sides:

4x = 16

Divide both sides by 4:

x = 4

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The difference of the two means is not significant, so the null hypothesis must be rejected.


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(2n-9)-(-2.4n+4) Please help!
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Answer: =4.4n-13

Step-by-step explanation:

Let's simplify step-by-step.

2n−9−(−2.4n+4)

Distribute the Negative Sign:

=2n−9+−1(−2.4n+4)

=2n+−9+−1(−2.4n)+(−1)(4)

=2n+−9+2.4n+−4

Combine Like Terms:

=2n+−9+2.4n+−4

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A hot air balloon goes 350 feet into the air and drifts to the east for 600 feet. How far is the balloon away from its launch po
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950 feet

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3 years ago
Find cos(2*ABC) 100POINTS
juin [17]

Answer:

-\dfrac{7}{25}

Step-by-step explanation:

<u>Trigonometric Identities</u>

\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

  • \cos(\angle ABC)=\dfrac{3}{5}
  • \sin(\angle ABC)=\dfrac{4}{5}

Therefore, using the trig identities and ratios:

\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}

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2 years ago
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