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antoniya [11.8K]
3 years ago
7

A 10-foot ladder is leaning against a wall. The base of the ladder is 2 feet from the base of the building. How far up the build

ing does the ladder fall?
Mathematics
1 answer:
Svetlanka [38]3 years ago
5 0
We can see on this graph that the triangle has legs of x and 6 with a hypotenuse of 10 and we can use Pythagoreans theorem to find the unknown side.
Pythagoreans theorem: a^2+b^2=c^2, where a and b are the legs of the triangle and c, is the hypotenuse 

x^2+6^2=10^2 Plugin a=x, b=6, and c=10. Now let us solve for x 

x^2+36=100 Square each individual term 

x^2+=100+36 Subtract 36 from both sides 

x^2=64 Combine like terms 

sqrt(2)=sqrt(64) Take the square root of both sides 

x = 8 Simplify the square root 

So our answer is x = 8 
The ladder touches the 8 feet mark on the wall.
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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

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The steepest part of any walking trail in Craters of the Moon National Monument rises 9 yards vertically for every 50 yards trav
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Answer:

4.5

Step-by-step explanation:

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Which of the following expressions are equivalent?
Anni [7]
The answer is D ................... D.................. D
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Find the product of 3 1/5 and 1 2/7. Express your answer in simplest form.
nydimaria [60]

Answer:

4\frac{1}{56}

Step-by-step explanation:

3\frac{1}{8} * 1 \frac{2}{7} = 4\frac{1}{56}

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3 years ago
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Explain how you have solved for x and show your workings out. I will award the brainliest 25 points!
Brums [2.3K]

Answer:

The value of x is 10º.

Step-by-step explanation:

From Euclidean Geometry we remember that the sum of internal angles within a triangle equals to 180º. We present the resulting triangle after applying some geometric handling in the image attached below. Then, the triangle satisfies the following equation:

100^{\circ} + (80^{\circ}-5\cdot x)+(4\cdot x +10^{\circ}) = 180^{\circ} (1)

190^{\circ} - x = 180^{\circ}

x = 10^{\circ}

The value of x is 10º.

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3 years ago
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