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chubhunter [2.5K]
3 years ago
6

How much work in foot-pound is required to move a block 10 feet if the force is 200 lb

Mathematics
1 answer:
KatRina [158]3 years ago
4 0

             Work = (force) x (distance)

                       = (200 lb) x (10 ft)  =  2,000 ft-lbs .

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April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3
vlada-n [284]

Answer:

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = 5\frac{1}{4}\ hrs

5\frac{1}{4}\ hrs can be Rewritten as \frac{21}{4}\ hrs

Number of Hours Carl worked on Math project = \frac{21}{4}\ hrs

Number of Hours Sonia worked on Math project = 6\frac{1}{2}\ hrs

6\frac{1}{2}\ hrs can be rewritten as \frac{13}{2}\ hrs

Number of Hours Sonia worked on Math project = \frac{13}{2}\ hrs

Number of Hours Tony worked on Math project = 5\frac{2}{3}\ hrs

5\frac{2}{3}\ hrs can be rewritten as \frac{17}{3}\ hrs.

Number of Hours Tony worked on Math project = \frac{17}{3}\ hrs.

Now Given:

April worked 1\frac{1}{2} times as long on her math project as did Carl.

1\frac{1}{2}  can be Rewritten as \frac{3}{2}

Number of Hours April worked on math project = \frac{3}{2} \times Number of Hours Carl worked on Math project

Number of Hours April worked on math project = \frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs

Also Given:

Debbie worked 1\frac{1}{4} times as long as Sonia.

1\frac{1}{4}  can be Rewritten as \frac{5}{4}.

Number of Hours Debbie worked on math project = \frac{5}{4} \times Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = \frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs

Also Given:

Richard worked 1\frac{3}{8} times as long as tony.

1\frac{3}{8} can be Rewritten as \frac{11}{8}

Number of Hours Richard worked on math project = \frac{11}{8} \times Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = \frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs

Hence We will match each student with number of hours she worked.

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

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 Im pretty sure you just multiply 12 by 108, in that case there would be 2160 rows
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What are the factors of 17
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1 and 17 are the only factors.
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**Solve by Elimination<br> 3 + 2x - y = 0<br> -3-7y = 10x<br> 7) x =<br> 8) y =
timofeeve [1]
Answer:

Explanation:

3 + 2x - y = 0
Or 2x - y = -3 (1)

-3-7y = 10x
Or -10x - 7y = 3 (2)

2x - y = -3 (1)
-10x - 7y = 3 (2)
———————

Multiply 5 to (1)

5(2x - y = -3)
-10x - 7y = 3
———————
10x - 5y = -15
-10x - 7y = 3
———————
-12y = -12
y = -12/-12
y = 1

Plug y value in (1)

2x - 1 = -3
2x = -3 + 1
2x = -2
x = -2/2 = -1

Therefore, x = -1 and y = 1
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2 years ago
Algebra Please help<br> Simplify (2√5+3√7)^2
GREYUIT [131]
4√5+9<span>√7

Hope this helps!</span>
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