Find the exact value of x in the triangle below

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

So this is a special triangle. The 30-60-90 triangle rule states that if the short leg is y, then the hypotenuse is 2y and the long leg is y√3.

In this case, the short leg is 5√3 since that times √3 makes 15.

Now with the hypotenuse, just multiply 5√3 with 2, and your answer should be 10√3, or C.

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Find the product (x+y)(6x-8y)

QveST [7]

Use the distributive property and collect terms.
= x·6x -x·8y +y·6x -y·8y
= 6x² -2xy -8y²

4 0

4 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$