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4 years ago

11

What forms do the equations of motion for an idealized string take if either (a) the linear density varies with position, or (b)

the string hangs vertically supported only at the upper end?

1 answer:

stepan [7]4 years ago

7 0

Answer:

a) the varying densities are to be considered with respect to position

Step-by-step explanation:

the complete answer is found in the attachment

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IgorC [24]

Answer:

g(x) is shifted 4 units left and 6 units down from f(x).

Step-by-step explanation:

The parent function is:

f(x).

The child function is:

$g(x) = f(x+4) - 6$

Transformation 1:

$g(x) = f(x+4)$

Shifting a function f(x) a units to the left is finding f(x + a). So g(x) = f(x + 4) is f(x) shifted 4 units to the left.

Transformation 2:

$g(x) = f(x+4) - 6$

Subtracting a function f(x) by a constant a is the same as shifting the function a units down. So subtracting by 6 is shifting the function 6 units down. Thus, the correct answer is:

g(x) is shifted 4 units left and 6 units down from f(x).

5 0

3 years ago

Find the area of the trapezoid.

Firlakuza [10]

Answer:

141.12

Step-by-step explanation:

(12.2+10.2)/2*12.6=141.12

7 0

3 years ago

A loan of$8000 was repaid in 24 equal monthly installments of $400 . The rate of interest on the loan was

Lesechka [4]

Answer:

1%

Step-by-step explanation:

24(400) = 9600 = total amt paid

9600 - 8000 = 1600 = amt of interest paid

I = PRT T = 2 yr

1600 = 8000(2)R

1600 = 16000R

1600/16000 = 1/10 = .01 = 1%

8 0

3 years ago

Please help me ASAP!Please this is due soon and i don't have all day.

Helga [31]

<u><em>Answer:</em></u>

Section 1: 5.59%

Section 2: -2.16%

Section 3: -10.82%

Section 4: -6.54%

Section 5: -2.16%

Section 6: 7.86%

<u><em>Explanation:</em></u>

<u>The formula to calculate the percent change in mass is given as:</u>

$PercentChange = \frac{ChangeInMass}{StartingMass}*100=\frac{NewMass - OldMass}{OldMass}*100$

<u>1- Section 1:</u>

Old mass = 7.83 g and New mass = 8.268 g

$PercentChange = \frac{8.268-7.83}{7.83}*100 =$ 5.59%

<u>2- Section 2:</u>

Old mass = 7.4 g and New mass = 7.24 g

$PercentChange = \frac{7.24-7.4}{7.4}*100 =$ -2.16%

<u>3- Section 3:</u>

Old mass = 7.3 g and New mass = 6.51 g

$PercentChange = \frac{6.51-7.3}{7.3}*100 =$ -10.82%

<u>4- Section 4:</u>

Old mass = 7.49 g and New mass = 7.0 g

$PercentChange = \frac{7.0-7.49}{7.49}*100 =$ -6.54%

<u>5- Section 5:</u>

Old mass = 7.4 g and New mass = 7.24 g

$PercentChange = \frac{7.24-7.4}{7.7.4}*100 =$ -2.16%

<u>6- Section 6:</u>

Old mass = 7.89 g and New mass = 8.51 g

$PercentChange = \frac{8.51-7.89}{7.89}*100 =$ 7.86%

Hope this helps :)

6 0

3 years ago

Someone please help me on this one

irakobra [83]

12lb=$27
14lb=$x

$\frac{12}{27} = \frac{14}{x}$

cross multiply

12x=378
divide both sides by 12
x=31.5= $31.50

Answer=$31.50

8 0

3 years ago