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alexdok [17]
3 years ago
13

Find the solution set: (x−3)(x+3)=8x

Mathematics
1 answer:
AVprozaik [17]3 years ago
6 0

Answer:

You move all the terms to the left of the equation and set equal to zero. Then set each factor equal to zero.

X = 9, -1

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Show that √(1-cos A/1+cos A) =cosec A - cot A​
Gennadij [26K]

Hi there!

\sqrt{\frac{1-cosA}{1+cosA}} =

We can begin by multiplying by its conjugate:

\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}}  = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =

Simplify using the identity:

1 - cos^2A = sin^2A

\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =

Take the square root of the expression:

{\frac{sinA}{1+cosA} =

Multiply again by the conjugate to get a SINGLE term in the denominator:

{\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\

Simplify:

{\frac{sinA(1-cosA)}{1-cos^2A} =

Use the above trig identity one more:

{\frac{sinA(1-cosA)}{sin^2A} =

Cancel out sinA:

{\frac{(1-cosA)}{sinA}  =

Split the fraction into two:

{\frac{1}{sinA}   - \frac{cosA}{sinA} =

Recall:

1/sinA = cscA\\\\cosA/sinA = cotA

Simplify:

\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}

5 0
2 years ago
G(n) = n + 1 Find g(g(n))
mars1129 [50]

Answer:

the answer would be:

n=0

Hope that helps! :)

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Find the limit (if it exists). (If an answer does not exist, enter DNE.) lim x→(−1/2)+ 6x2 + x − 1 4x2 − 4x − 3
Evgesh-ka [11]
You might be able to do direct substitution. 
5 0
2 years ago
please help 100 points!!! will give crown to best answer if f(x) = 3x^2 -2x + 4 and g (x) = 5x ^2 + 6x -8 find (f-g) (x)
Alexxandr [17]
(f-g)(x) will be 3x^2 -2x+4 -(5x^2+6×-8)
distribute the -sign, gives us
3x^2-2x+4-5x^2-6x+8,
now combine like terms and should get,
-2x^2-8x+12,
final answer,
(f-g)(x)= -2x^2-8x+1×,
hope it helped,
good luck
8 0
3 years ago
Read 2 more answers
How do you work this problem 2.5÷ 4
lord [1]

Using a calculator can help. You get 0.625, or 5/8.

8 0
3 years ago
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