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borishaifa [10]
3 years ago
8

Last questions; I NEED HELP!

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0
Given that cos Ф = 5/6 I know that one side is 5 and the hypotenuse is 6 so the other side is:
a^2 = c^2 - b^2
a^2 = 6^2 - 5^2
a^2 = 36 - 25 = 11
a = \sqrt{11}

Now I know that the sin is \frac{ \sqrt{11} }{6} since sin is defined as opposite/hypotenuse

the second one uses a form of C = pi * d we are going to add the sector part to it.
C = 6 and the central angle is 4 degrees we can set up like this:
6 =  \frac{4}{360}* \pi *d
\frac{6 * 360}{4 *  \pi } = d

d = 171.974
r = 171.94/2 =85.987
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Santos went on a bike ride of 90 miles. He realized that if he had gone 24 mph faster, he would have
Gnoma [55]
<h3>Answer:  6 mph</h3>

========================================

Explanation:

distance = rate*time

90 = r*t

where r is the slower speed and t is the time it takes when going that slower speed.

If r is bumped up 24 mph faster, to r+24, then Santos takes t-12 hours to get there. The second equation is

90 = (r+24)(t-12)

-------------------------------------

We can solve the first equation for r to get r = 90/t

Then plug this into the second equation and do a bit of algebra

90 = (r+24)(t-12)

90 = (90/t+24)*(t-12)

90 = 90 - 1080/t + 24t - 288

90t = 90t - 1080 + 24t^2 - 288t

0 = 90t - 1080 + 24t^2 - 288t-90t

0 = 24t^2 - 288t - 1080

24t^2 - 288t - 1080 = 0

If you apply the quadratic formula, then you should get the two solutions t = -3 and t = 15. Due to time constraints, I'll skip these steps.

We'll ignore the negative t value. It makes no sense to have a negative time.

So we focus on t = 15 as the only solution.

-------------------------------------

If t = 15, then,

r = 90/t

r = 90/15

r = 6

Santos rode his bike at <u>6 mph</u> at first. Going this speed means he takes 15 hours.

If he rode 24 mph faster, at 6+24 = 30 mph, then he would ride for 15-12 = 3 hours instead. Note that 90/30 = 3.

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