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3 years ago

Pls help me on #3. I been having trouble on this I’m not so good at math

Mathematics

1 answer:

natali 33 [55]3 years ago

3 0

Answer:

I think the answer is 5 if I'm right

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You make $10.50 an hour at your part-time job and Andy makes $9.75 an hour at his full-time job. If you work a full week of 40 h

Marianna [84]

The awnser is $420.00 (10.50 X 40 = 4200)

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2 years ago

Two cars are traveling in the same direction. The first car is going 40 mi/h and the second car is going 55 mi/h. The first car

dexar [7]

Answer:

The equation you are asking for is

40*(t+3) = 55*t,

where t is the time counted after the second car started.

The equation says that the distance covered by each car from the starting point to the catching point is the same for both cars.

Step-by-step explanation:

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3 years ago

Combine like terms to create an equivalent expression -4/7p+(-2/7p)+1/7

STatiana [176]

-6/7p+1/7 um Idk what you wanted me to do but I added -4/7p with -2/7p

7 0

3 years ago

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What is the sum of a 7-term geometric series if the first term is −11, the last term is −171,875, and the common ratio is −5?

mr_godi [17]

The answer is -143231
hope this helps:]

4 0

3 years ago

Read 2 more answers

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

maxonik [38]

Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

$F = \dfrac{9}{6^x}$

This force is not a constant force. For a non-constant force, the work done, <em>W</em>, is

$W = \int\limits^{x_2}_{x_1} {F(x)} \, dx$

with $x_1$ and $x_2$ the initial and final displacements respectively.

From the question, $x_1 =0$ and $x_2 = 12$ .

Then

$W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx$

Evaluating the indefinite integral,

$\int\limits \dfrac{9}{6^x} \, dx =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx$

From the rules of integration,

$\int\limits a^x\, dx = \dfrac{a^x}{\ln a}$

$9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x$

Returning the limits,

$\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}$

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3 years ago