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natta225 [31]
3 years ago
9

A building is 2 ft from a 12 ft fence that surrounds the property. A worker wants to wash a window in the building 17 ft from th

e ground. He plans to place a ladder over the fence so it rests against the building. He decides he should place the ladder 8 ft from the fence for stability. To the nearest tenth of a foot, how long of a ladder will he need?

Mathematics
2 answers:
Marta_Voda [28]3 years ago
3 0
17^2+10^2=x^2
x=19.7
2+8=10

vovikov84 [41]3 years ago
3 0
The ladder would have to be about 19.7 feet tall

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pickupchik [31]
<em>46 - 40 = 6</em>

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<em>one hour = $17.90</em>

<em>overtime rate = $17.90 x 1.5 = $</em><span><em>26.85</em>

His overtime premium pay is $26.85

</span><em>$26.85 x 6 = $161.10 </em>

He is paid $161.10 for 6 hours


Answer: He is paid $161.10 for the workweek.
7 0
3 years ago
Calculate the ROI, given the following: Andrew invested $20,000 in mutual funds and received a sum of $35,000 at the end of the
Dennis_Churaev [7]
We are given with an initial deposit of  $20,000 and a future worth of <span>$35,000. In this case, we are asked for the return of income (ROI) of the investment. in this case, we assume the number of years equal to 1. hence, 

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3 0
3 years ago
Read 2 more answers
3(2x+1)-2(x+1) is the same as
Dahasolnce [82]

Answer:

4x + 1 and 6x - 2x - 2 + 3

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and another being 4x + 1

hope this helps!

5 0
2 years ago
Solve y + -7 = 13 ?
ratelena [41]
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3 0
3 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
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