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1 answer:

Lapatulllka [165]3 years ago

4 0

The answer for this problem is D

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Find the angle in degrees

BlackZzzverrR [31]

Answer:

90 degrees

Step-by-step explanation:

its a 45-45-90 triangle

3 0

3 years ago

A ladder 16 feet long is leaning against the wall of a tall building. The base of the ladder is moving away from the wall at a r

svet-max [94.6K]

Answer:

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Step-by-step explanation:

Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem, $h^2 + b^2 = 16^2 = 256 \dotsc\;(1)$ because it is a right angle triangle.

It is given that $\frac{db}{dt} = 1$

Now differentiate (1) with respect to t (time) :

$\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}$

$\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}$

The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.

b. From the diagram, infer that $16 \sin{\theta} = b$ . When b = 8, then $\theta = \arcsin{0.5} = \ang{30}$ .

Differentiate the above equation w.r.t t

$\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}$