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miskamm [114]
3 years ago
8

Pls Help I Need This Answer ASAP ​

Mathematics
2 answers:
harkovskaia [24]3 years ago
4 0

Answer:

ℎ = − 1 6 /7

Step-by-step explanation:

Step 1: Combine like terms.

ℎ − 8 ℎ + 1 9 ℎ − 5 ℎ = − 1 6

7 ℎ = − 1 6

Step 2: Divide both sides of the equation by the same term.

7 ℎ = − 1 6

7ℎ/7=−16/7

Step 3: Simplify

ℎ=−16/ 7 = Answer

Alex777 [14]3 years ago
3 0

Answer:

h = -2

Step-by-step explanation:

1. Simplify the left hand side: 8h = -16

2. DIvide both sides by 8: h = -16 ÷ 8 = -2

Hope this helps!

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Two figures have a similarity ratio of 5:8. If the volume of the smaller figure is 875 mm, what is the
Akimi4 [234]

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Step-by-step explanation:

3 0
3 years ago
Find the domain and range of the relation and determine whether it is a function.
Anna [14]
The correct answer to this question is this "domain: x > 1; range: y > 0; Yes, it is a function."

Based from the graph, <span>it has a vertical asymptote at x = 1, so domain is x > 1 </span><span>and since it has horizontal asymptote at y=0, its range is y > 0. So this concludes us to have a domain of x > 1 and a range of y > 0.</span>
7 0
3 years ago
When the reciprocal of the larger of two consecutive even integers is subtracted from 4 times the reciprocal of the smaller, the
yawa3891 [41]

Answer:  No solution

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

1st number: x

2nd number: x + 2

  • reciprocal of the larger is \frac{1}{x+2}
  • 4 times the reciprocal of the smaller is 4(\frac{1}{x})
  • reciprocal of the larger of two consecutive even integers is subtracted from 4 times the reciprocal of the smaller is 4(\frac{1}{x}) - \frac{1}{x+2}
  • the result is 56: 4(\frac{1}{x}) - \frac{1}{x+2} = 56

Solve the equation:

4(\frac{1}{x}) - \frac{1}{x+2} = 56

\frac{4(x(x + 2))}{x} - \frac{1(x(x + 2))}{x+2} = 56(x(x + 2))

4x + 8 - x = 56x² + 112x

     3x + 8 = 56x² + 112x

             0 = 56x² + 109x - 8

Use the quadratic formula to solve:

x = \frac{-109+/-11\sqrt{113}}{112}

The result is NOT AN INTEGER so there is NO SOLUTION

8 0
3 years ago
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