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y^(-2)+2y^(-1)-15=0

what is a possible value of y

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Asked by C from Venice, FL
20

y-+ 2y-1 - 15 = 0

This might be easier with a change of variables; let's replace y-1 with a, which makes y-1 = a and  y-2 =   (y-1)2 = a2 and the expression becomes...

a2 + 2a - 15 = 0 which factors to:

(a + 5)(a - 3) = 0

Either a + 5 = 0 or a - 3 = 0 so,

a = -5 or a = 3

Now change a back to y-1:

y-1 = -5 or y-1 = 3

Remember that y-1 = 1/y and you should be able to solve it from here. Make sure to check that the values for y are valid, by putting them back into the original equation and checking it equals zero.

find the radius and center of the cirles

x2+y2 -4x-6y-12=0

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Asked by Moustafa from New York, NY
10

First you have to complete the square with both the y and the x. Start by grouping together the x's and y's, and you might as well add 12 to both sides to get:

(x2 - 4x) + (y2 - 6y) = 12

Now complete the square, by dividing the -4 in front of the x by 2, and divide the -6 in front of the y by 2:

(x - 2)2 + (y - 3)2 = 12 + 4 + 9

When we completed the square, we added a (-2)2 and a (-3)2 to the left, so we have to add them to the right (the +4 and +9)

(x - 2)2 + (y - 3)2 = 25

From here you can easily get the center and radius. Use the standard form of a circle:

(x - h)2 + (y - k)2 = r2

The center is (h, k) and the radius is r.

I need help with a function in Algebra which is posted below

How do you slove this problem.. f(4)f(x)= 4-2x^2

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Asked by Colleen from Jacksonville, FL
10

I think you have the function f(x)= 4 - 2x2 and you want to evaluate it for when x = 4. If so, you're finding f(4), so you'd plug in 4 as x, like so:

f(4) = 4 - 2*22 = ?

Whats X + 3 > 2 ?

Inequalities

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Asked by Agueda from Palo Alto, CA
10

With inequalities, you mostly treat them the same as an equality, except when multiplying or dividing both sides by a negative number (that changes the direction of the inequality sign).

In this case, treat x + 3 > 2 the same way you'd treat x + 3 = 2. We want to get all the numbers on one side, and leave the x variable on the other side. First start by subtracting 3 from both sides

x + 3 > 2

x + 3 - 3 > 2 - 3

x > -1

And you're done. The solution here, is that x must be greater than -1.

solve (x+3)(x+6)=4

I have to solve (x+3)(x+6)=4 setting the equation to zero.  Im not sure what to do once I move the 4 to the left to make it (x+3)(x+6)-4=0

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Asked by Suhaili from South El Monte, CA
10

First try FOILing the two sets of parentheses:

(x+3)(x+6)-4=0

x2 + 3x + 6x + 18 - 4 = 0

x2 + 9x + 14 = 0

Now do you know how to factor this? Look for two numbers that multiply to equal 14, and add up to 9, then you can put the factors into the parentheses:

(x + _)(x + _) = 0

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