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zmey [24]
3 years ago
5

Solve kx + 12 = 3kx for x.

Mathematics
2 answers:
OLga [1]3 years ago
4 0

kx+12=3kx \\12=2kx \\x=\frac{12}{2k} \\x=\frac{6}{k}

Hope this helps.

KengaRu [80]3 years ago
4 0

Answer: x=6/k

Step-by-step explanation:

kx + 12 = 3kx and subtract k from both sides

12=3kx-kx then simplify

12=2kx, now divide both sides by two

12/2=kx simplify again

6=kx divide next

6/k=x finally, switch sides to get x=6/k

Hope this helps, have a BLESSED and wonderful day, as well as a safe one!

-Cutiepatutie ☺❀❤

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BlackZzzverrR [31]

Step 1: Simplify both sides of the equation.

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17+−2n=9

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3 years ago
What is the answer to this
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Answer:

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Ugo [173]
Hello Friend,here is the solution for your question


<span>so the given function is </span>
y= √(-2cos²x+3cosx-1)
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i.e = √[-2(cosx-3/4)²-9/16+1/2]
i.e. = √[-2(cos-3/4)²-1/16]
i.e. = √[1/8-3(cosx=3/4)²]-----------(1)

Now here  in this equation is this quantity :-
<span>(cosx=3/4)²----------------(2)   is to it's minimum value then the whole equation </span>
<span>i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa </span>


And we know that cosx-3/4 will be minimum if cosx=3/4
<span>therefore put this in (1) we get </span>
(cosx=3/4)²=0    [ cosx=3/4]
<span>hence the minimum value of the quantity (cosx=3/4)² is 0 </span>

<span>put this in equation (1) </span>
we get ,
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   =√[1/8-3(0)]        [ because minimum value of of the quantity (cosx=3/4)² is 0 ]
     =√1/8
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<span>this is the maximum value now to find the minimum value </span>

<span>since this is function of root so the value of y will always be ≥0 </span>

<span>hence the minimum value of the function y is 0 </span>


<span>Therefore, the range of function </span>y is [0,1/(2√2)]


__Well,I have explained explained each and every step,do tell me if you don't understand any step._

8 0
3 years ago
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