Your hands are slow,

Mathematics

2 answers:

solmaris [256]3 years ago

4 0

Are ya trying to make me thirsty right now

aivan3 [116]3 years ago

3 0

Answer:

Step-by-step explanation:

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A car is approaching a building. When it is 300 ft away, the angle of depression from the top of the building to the car is 42°.

labwork [276]

Answer: 333.2ft

Step-by-step explanation:

We can draw a triangle rectangle, where the distance between the base of the building and the car is one cathetus (300ft)

The height of the building is the other cathetus.

Now, we know that the angle of depression from the top of the building to the car is 42°

This angle is measured from a perpendicular line in from the building, so the "top angle" of the triangle rectangle will be:

90° - 42° = 48°

if we steep on this angle, the 300ft cathetus is the opposite cathetus and the height of the building is the adjacent cathetus.

Here we can use the relation:

Tan(A) = Opposite cathetus/adjacent cathetus:

Then:

Tan(48°) = H/300ft

Tan(48°)*300ft = H = 333.2ft

7 0

3 years ago

Read 2 more answers

Evaluate the limit, if it exists. (if an answer does not exist, enter dne.) lim h→0 (5 + h)−1 − 5−1 h

34kurt

$\displaystyle\lim_{h\to0}\frac{(5+h)^{-1}-5^{-1}}h=\lim_{h\to0}\frac{\frac5{5(5+h)}-\frac{5+h}{5(5+h)}}h$
$\displaystyle=-\lim_{h\to0}\frac h{5(5+h)h}$
$\displaystyle=-\lim_{h\to0}\frac1{5(5+h)}=-\frac1{25}$

Alternatively, recall that if $f(x)=\dfrac1x$ , then $f'(x)=-\dfrac1{x^2}$ , and so

$f'(5)=\displaystyle\lim_{x\to5}\frac{\frac1x-\frac15}{x-5}$

Take $h=x-5$ , so that $x=h+5$ , and we have the original limit. So the limit is equivalent to the value of $f'(5)$ , i.e.

$f'(5)=-\dfrac1{5^2}=-\dfrac1{25}$