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3 years ago

Solve the inequality -2n > -6 A. n>3 B. n>-3 C. n<3 D. n=3

Mathematics

1 answer:

Natasha2012 [34]3 years ago

5 0

The answer is C ... or you could say n<3

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Solve: 2x-3 = 5(x + 6)

Slav-nsk [51]

Answer:

2x - 3 = 5(x + 6)

2x - 3 = 5x + 30

-33 = 3x

x = -11

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3 years ago

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Solve for y<br>7y + 9 - 10y = -18<br>What is y simplify your answer as much as possible.

Tpy6a [65]

Y = 9

7y - 10y = -3y
bring down the 9 = -18 so

-3y + 9 = -18

9 - 9 = 0
- 18 - 9 = - 27

- 3y ÷ - 3 = y
- 27 ÷ -3 = 9

therefore y =9

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3 years ago

Lilly takes a train each day to work that averages 35 miles per hour. On her way home, her train ride follows the same path and

Klio2033 [76]

Answer: $2.5=\frac{n}{35}+\frac{n}{45}$

Step-by-step explanation:

1. To find the equation asked, you must add the times:

$t_{total}=t_1+t_2$

Where $t_1$ is the time takes Lilly in her path to work and $t_2$ is the time takes Lilly in her path to home.

2. You must use the following formula of speed and solve for the time

$V=\frac{n}{t}$

Where n is the distance, V is the speed and t is the time.

3. You know that Lilly takes a train each day to work that averages 35 miles per hour, then, you can write the following expression:

$35=\frac{n}{t_1}\\t_1=\frac{n}{35}$

4. And her train ride follows the same path at 45 miles per hour:

$45=\frac{n}{t_2}\\t_2=\frac{n}{45}$

5. Then, you obtain the following equation:

$2.5=\frac{n}{35}+\frac{n}{45}$

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3 years ago

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When Joseph first starts working at a grocery store, his hourly rate is $10. For each year he works at the grocery store, his ho

stiks02 [169]

Answer: R(t) = 10 + 0.50t

Step-by-step explanation:

From the question, we've been informed that Joseph's hourly rate is $10 and that for every year he works at the grocery store, his hourly rate increases by $0.50.

Therefore, the function will be

R(t) = 10 + 0.50t

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3 years ago

Simplify this problem

ikadub [295]

Answer:

$\boxed{ \frac{ \sqrt[3]{ {x}^{11} } }{4} }$

Step-by-step explanation:

$= > \frac{ {x}^{4} }{ \sqrt[3]{64x} } \\ \\ = > \frac{ {x}^{4} }{ {(64x)}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4} }{ ({64}^{ \frac{1}{3} } )\times ({x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({( {4}^{3} )}^{ \frac{1}{3} }) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({4}^{ \cancel{3} \times \frac{1}{ \cancel{3}} } ) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{4 {x}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4 - \frac{1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{12 - 1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{11}{3} } }{4} \\ \\ = > \frac{ \sqrt[3]{ {x}^{11} } }{4}$

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3 years ago