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3 years ago

10

John drove from station A to station B a distance of 224 miles. On his way back, he increased his speed by 10 mph. If the journe

y back took him 24 minutes less, what was his original speed?

1 answer:

meriva3 years ago

5 0

Answer: 70 mph

Step-by-step explanation:

224/x =224/x+10 +24/60

Solve equation

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Read 2 more answers

Given cos theta=4/7 and csc theta<0, find sin theta and tan theta.

Ksenya-84 [330]

ANSWER

The correct answer is C

$\sin ( \theta) = - \frac{ \sqrt{33} }{7} , \tan ( \theta) = - \frac{ \sqrt{33} }{ 4 }$

EXPLANATION

It was given that,

$\cos( \theta) = \frac{4}{7}$

and

$\csc( \theta) < 0$

This means that,

$\theta$
is in the fourth quadrant.

We use the identity,

$\cos ^{2} ( \theta) + \sin ^{2} ( \theta) = 1$

This implies that,

$( { \frac{4}{7} })^{2} + \sin ^{2} ( \theta) = 1$

${ \frac{16}{49} } + \sin ^{2} ( \theta) = 1$

$\sin ^{2} ( \theta) = 1 - { \frac{16}{49} }$

$\sin ^{2} ( \theta) = { \frac{33}{49} }$

$\sin ( \theta) = \pm \sqrt{{ \frac{33}{49} }}$

$\sin ( \theta) = \pm \frac{ \sqrt{33} }{7}$

But

$\csc( \theta) < 0$

This implies that,

$\sin ( \theta) = - \frac{ \sqrt{33} }{7}$

$\tan ( \theta) = \frac{ \sin( \theta) }{ \cos( \theta) }$
$\tan ( \theta) = \frac{ - \frac{ \sqrt{33} }{7} }{ \frac{4}{7} }$

$\tan ( \theta) = - \frac{ \sqrt{33} }{ 4 }$

7 0

3 years ago