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3 years ago

Using the diagram below, what is the horizontal distance between the control tower and the plane?

Mathematics

1 answer:

Natali [406]3 years ago

6 0

Answer:

x = 664.093 m

Step-by-step explanation:

tan∅ = opposite over adjacent

Since we are trying to find the horizontal distance, that means we are dealing with the bottom length and not the hypotenuse (the diagonal length):

Our theta (∅) = 52°

Our opposite = 850 m

Our adjacent = x m

tan52° = 850/x

xtan52° = 850

x = 850/tan52°

x = 664.093 m

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student expanded an expression, as shown. Is the student's work correct? Explain why or why not. −6 ( 4x − 2 13 ) −6(4x) + 6 ( −

sleet_krkn [62]

Answer:

No, the student's work is not correct.

Step-by-step explanation:

Given : Student expanded an expression, as shown.

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The expansion of student is not correct.

Follow the below steps to get correct solution and student mistake,

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The student was mistaken in step 2 in solving the sign.

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3 years ago

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nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

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2 years ago

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marshall27 [118]

Answer:

Step-by-step explanation:

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