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nikdorinn [45]
3 years ago
15

Please help me!!! show work

Mathematics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:x=8 or x=-2

Step-by-step explanation:

log4(x-6) + log4(x)=2

log4(x(x-6))=2

4^2=x(x-6)

4x4=x(x-6)

16=x^2-6x

Arrange the quadratic equation

x^2-6x-16=0

x^2+2x-8x-16=0

x(x+2)-8(x+2)=0

(x-8)(x+2)=0

x-8=0. x+2=0

x=8. x=-2

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So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (0,2), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=0 and y1=2.

Also, let's call the second point you gave, (-4,0), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=-4 and y2=0.

Now, just plug the numbers into the formula for m above, like this:

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y=1/2x+b

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(-4,0). When x of the line is -4, y of the line must be 0.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=1/2x+b. b is what we want, the 1/2 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (0,2) and (-4,0).

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