Please help me ! !!!

Mathematics

1 answer:

Kruka [31]2 years ago

8 0

3 that’s the answer your welcome

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What is the measure of PRQ?

natima [27]

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3 years ago

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Lf x+1/x=12 then find the value of x3+1/x3

Tresset [83]

Answer:

Explanation:

First we find what x is:
x + 1/x = 12
x + 1 = 12x
1 = 12x - x
1 = 11x
1/11 = x
Or x = 1/11

Plug x value in x^3 + 1/x^3

(1/11)^3 + 1/(1/11)^3
= (1^3/11^3)+ 1/(1^3/11^3)
= (1/1331 + 1)/1/1331
= (1/1331 + 1331/1331)/1/1331
= 1332/1331 x 1331/1
= 1332/1
= 1332

Therefore, x^3 + 1/x^3 = 1332

4 0

3 years ago

Express 192 as a product of its prime factors.<br> write the prime factors in ascending order

kodGreya [7K]

Hi,
That would be 192= 2x2x2x2x2x2x3 = (2^6)x3

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3 years ago

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Sin(x+pi/4)-sin(x-pi/4)=1 solve the equation

evablogger [386]

Sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

$sin(x+ \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} \\ sin(x- \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} \\ \\ \\sin(x+ \frac{ \pi }{4} )-sin(x- \frac{ \pi }{4} ) =1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -(sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} )=1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\ cos(x)sin\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\$
$2cos(x)sin\frac{ \pi }{4} =1 \\ sin\frac{ \pi }{4} = \frac{ \sqrt{2} }{2} \\ 2cos(x) \frac{ \sqrt{2} }{2} =1 \\ 2 \cdot \frac{ \sqrt{2} }{2}cos(x) =1 \\ \sqrt{2} cos(x)=1 \\$
$cos(x)= \frac{1}{ \sqrt{2} } \\ x=\pm arccos \frac{1}{ \sqrt{2} }+2 \pi k , k \in Z \\ x=\pm \frac{ \pi }{4} +2 \pi k , k \in Z$

6 0

3 years ago