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klio [65]
3 years ago
15

Part A: Solve −np − 70 < 40 for n. Show your work. (4 points) Part B: Solve 4w − 7k = 28 for k. Show your work. (6 points

Mathematics
1 answer:
likoan [24]3 years ago
7 0
For part A, lets write the problem

-np - 70\ \textless \  40
 
add 70 to both sides

-np - 70 + 70 \ \textless \  40 + 70&#10;&#10;-np \ \textless \  110

we are going to reverse the inequality, to do this we are going to multiply both sides by -1

(-np ) ( -1 ) \ \textgreater \  110 ( -1 )

pn \ \textgreater \  -110

divide both sides by p

\frac{pn}{p}  \ \textgreater \  \frac{-110}{p}

our answer would be 

n \ \textgreater \    \frac{-110}{p}

lets go with part b, lets write the problem

4w - 7k = 28

subtract 4w from both sides

4w - 7k - 4w = 28 - 4w

-7k = 28 - 4w

divide both sides by -7

\frac{-7k}{-7}  =  \frac{28}{-7} - \frac{4w}{-7}

I few simplified it, we would be left with,

k =  -\frac{28 - 4w }{7}

I hope this helped!!

- Kiana 
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Harry has a penny collection with 160 pennies. He plans to increase the size of his collection by a constant term each year for
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Answer:

  • A: Harry plans to have approximately 350 pennies at the end of the second year

  • B: Harry plans to have approximately 1,200 pennies at the end of his five-year plan.

  • C: Harry plans to increase his collection by approximately 1,050 pennies over the five-year period.

  • D: Harry will add approximately 250 pennies to his collenction d during the fourth year, according to his plan.

  • E: The coordinates of the point are (3.5, 400)

  • F: Harry will need to add approximately 150 more pennies more than what he had planned by the end of the fourth year, to get back on the plan.

Explanation:

The first thing I did was to write the numbers to the grid and add the points labeled with the letter of the question with which they are related.

  • Every mark on the x-axis corresponds to 1 year.

  • Every mark of the y-axis corresponds to 200 pennies.

See the graph attached.

Question A:

The y-coordinate of the green point labeled A is about half way between 300 and 400, i.e. 350.

Then, Harry plans to have approximately 350 pennies at the end of the second year

Question B:

The green point labeled B has coordinates (5, 1200), meaning that Harry plans to have approximately 1,200 pennies at the end of his five-year plan.

Question C:

Harry started his collection with approximately 150 coins. It is the y-coordinate (the value when x = 0): (200 - 100)/2 = 150.

Thus, Harry plans to increase his collection by approximately 1,200 - 150 = 1,050 pennies over the five-year period.

Question D:

The number of peenies that Harry will add to his collection during the fourth year, according to his plan, is the difference between the number of pennies at the end of the fourth year and the end of the third year.

The green point labeled D has coordinates (4, 800) meaning that his plan is to have approximately 800 pennies.

To find the number of pennies that Harry plans to have at the end of the third year  read the y-coordinate that meets the curve when x = 3. That is approximately (600 + 500) / 2 = 550.

The difference is 800 - 550 = 250 pennies.

Therefore, Harry will add approximately 250 pennies to his collenction d during the fourth year, according to his plan.

Question E:

The blue point labeled E shows the position on the grid of the 400 pennies halfway throuhg the fourth year.

The coordinates of this point is (3.5, 400).

Question F:

According to the plan, the number of pennies at the end of the fourth year would be 800 (green point labeled D on the graph).

Then, Harry will need to add approximately 800 - 400 = 400 pennies.

According to the plan the number of coins at the end of the third year would be

But the plan was to add 800 - 550 = 250.

Thus, Harry will need to add 400 - 250 = 150 more pennies than he had planned, during the fourth year.

7 0
3 years ago
Read 2 more answers
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