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melamori03 [73]
3 years ago
14

A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the

bottom of the ladder? Round to the nearest tenth of a foot
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
6 0

Answer:

The wall is 10.5 foot far from the bottom of the ladder.

Step-by-step explanation:

Given:

Height of the ladder leaning against wall (Hypotenuse) = 20 foot.

Height from where the window is above from the ground (Leg1) = 17 feet.

To find the distance of the bottom of the ladder far from the wall (Leg2) = ?

Now, by using the pythagorean theorem:

Hypotenuse^{2} = (Leg1)^{2} + (Leg2)^{2}

20^{2} =17^{2}+(Leg2) ^{2}

400=289+(Leg2)^{2}

400-289=(Leg2)^{2}

111=(Leg2)^{2}

by squaring both sides

Leg2=10.535

10.535 foot rounded nearest tenth is 10.5 foot.

Therefore, the wall is 10.5 foot far from the bottom of the ladder.  

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Answer:

I would say 16=1.2n+(negative 8) because they are in the same order as it was spelled in words and not numbers that's how I know that it could be 16=1.2n+(negative 8).

Step-by-step explanation:

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iren [92.7K]

Answer:

y=2x + 2

0= 2x +2

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Step-by-step explanation:

2y = 4x +4

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3 years ago
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wlad13 [49]

Answer:

P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69

And we can round this to the nearest up integer and we got 110194.  

Step-by-step explanation:

The natural growth and decay model is given by:

\frac{dP}{dt}=kP   (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

\int \frac{dP}{P} =\int kdt

ln|P| =kt +c

And if we apply exponentials on both sides we got:

P= e^{kt} e^k

And we can assume e^k = P_o

And we have this model:

P(t) = P_o e^{kt}

And for this case we want to find P_o

By 1990 we have t=20 years since 1970 and we have this equation:

143000 = P_o e^{20k}

And we can solve for P_o like this:

P_o = \frac{143000}{e^{20k}}   (1)

By 2019 we have 49 years since 1970 the equation is given by:

98000 = P_o e^{49k}   (2)

And replacing P_o from equation (1) we got:

98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}  

We can divide both sides by 143000 we got:

\frac{98000}{143000} =0.685 = e^{29k}

And if we apply ln on both sides we got:

ln(0.685) = 29k

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69

And we can round this to the nearest up integer and we got 110194.  

7 0
3 years ago
a train moves at a constant speed. it can travel 150 miles in 2 1/2 hours. how long will it take for the train to travel 360 mil
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Answer:

It would equal 6 hours

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8 0
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I need help plz I don't get this problem
OlgaM077 [116]
To find the answer you need to write equivalent fractions with larger denominators. If you multiply each fraction by 3/3 then your equivalent fractions would be

3/5 × 3/3 = 9/15
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Now we have two fractions that are equivalent to the original fractions and we can easily see that 2 fractions between 9/15 and 12/15 would be...

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(10/15 can reduce to 2/3 so your answers could also be 2/3 and 11/15)
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