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pshichka [43]
2 years ago
7

Please help me, I have no idea what to do

Mathematics
2 answers:
DochEvi [55]2 years ago
6 0

Answer:

when x equals -3 y equals -23

when x equals 0, y equals -8

when x equals 3, y equals 7

when x equals 6, y equals 22

Step-by-step explanation:

Schach [20]2 years ago
4 0

Answer:

<h3>x=-3;Y=-23</h3><h3>x=0;Y=-8</h3><h3>x=3;Y=7</h3><h3>x=6;Y=22</h3>

Step-by-step explanation:

<h3>this question can be done by replacing the given value of X into the given equation </h3>

<h3>for the X value of -3 ,Y =5(-3)-8=-15-8=-23</h3>

<h3>for X=0;Y=5(0)-8 =0-8 = -8</h3>

<h3>for X=3;Y=5(3)-8 = 15-8 = 7</h3>

<h3>for X=6;Y=5(6)-8 = 30-8 = 2</h3>

<h3>I think its helpful </h3>
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(6x - 12)/3 + 4 = 18/x
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3 years ago
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jarptica [38.1K]
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real. 

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result. 

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
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5 0
3 years ago
What is 1-cos(6x)=___
FromTheMoon [43]

Answer:

2sin^2(3x)

Step-by-step explanation:

3 0
2 years ago
Algebra help? ♡ (25 pts)
natulia [17]
Yes the first choice is the right choice.

Remember that the 3rd root range is from -infinity to infinity.
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