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3 years ago

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Solve the literal equation for y; cy+3=6d-2y

2 answers:

victus00 [196]3 years ago

7 0

<span>cy+3=6d-2y
cy + 2y = 6d - 3
(c + 2)y = 6d - 3
y = (6d - 3)/(c + 2)</span>

densk [106]3 years ago

3 0

I will do my best to explain what is done, but feel free to ask if you have questions on what I did:
$cy+3=6d-2y$
bring all the "y"s to one side
$cy+3-3=6d-2y-3$
$cy+2y=6d-3-2y+2y$
this should be your result...
$cy+2y=6d-3$
now divide y out to isolate it:
$y(c+2)=6d-3$
then divide our c+2 so your y is alone
$\frac{y(c+2)}{c+2}= \frac{6d-3}{c+2}$
you should get this as your answer:
$y= \frac{6d-3}{c+2}$

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cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

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Rationalize, $\frac{7}{\sqrt{65} }$ x $\frac{\sqrt{65} }{\sqrt{65} }$ = $\frac{7\sqrt{65} }{65}$

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - $\frac{7\sqrt{65} }{65}$

cos θ = adj/hyp = $\frac{4}{\sqrt{65} }$

By rationalizing and knowing that cos θ is negative, cos θ = - $\frac{-4\sqrt{65} }{65}$

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sec θ = 1/cos θ = 1/ $\frac{4}{\sqrt{65} }$ = - $\frac{-\sqrt{65} }{4}$

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