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3 years ago

X^2+4x-1/(x+2)^2=0 solve for x

1 answer:

5 0

x^2+4x-1/(x+2)^2=0
x^2 + 4x - 1 = 0

$\frac{x^2+10x-7}{x^2+10x+25} =0\\x^2+10x-7=0\\x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ;
where a = 1, b = 4 and c = -1

$x=\frac{-4\pm\sqrt{4^2-(4\times1\times(-1)}}{2\times1}\\=\frac{-4\pm\sqrt{16+4}}{2}=\frac{-4\pm\sqrt{20}}{2}\\=\frac{-4\pm\sqrt{4\times5}}{2}=\frac{-4\pm2\sqrt{5}}{2}=-2\pm\sqrt{5}\\=-2+\sqrt{5}
 \ or \ -2-\sqrt{5}\\x=0.24 \ or \ x = -4.24$

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32 + 4 (2 + 4) X 6 = 8 + 4 =

Svetradugi [14.3K]

Answer:

176=12

Step-by-step explanation:

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2 years ago

Find the volume of the composite solid. Round your answer to the nearest tenth

Vesna [10]

Answer:

V = 1206.3 $ft^{3}$

Step-by-step explanation:

This shape is made up of a cylinder on the bottom and a cone on the top. We'll find the volumes of these shapes separately and then add them together.

Volume of a cylinder = area (of the base) x height

Substitute in the formula for the area of a circle.

V(cylinder) = $\pi$ $r^{2}$ x h

Substitute in the values for the radius (8) and height (4)

V(cylinder) = $\pi$ x $8^{2}$ x 4

Evaluate using a calculator

V(cylinder) = 804.2477

To the nearest tenth, V(cylinder) = 804.2

Volume of a cone = $\frac{\pi r^{2}h }{3}$ . This is the area of the circular part of the cone ( $\pi r^{2}$ ), multiplied by the height from the point to the base, all divided by 3.

Substitute in the values for the radius of the circle (8) and the height (6)

V(cone) = $\frac{\pi 8^{2} 6 }{3}$ (On the top it's $\pi$ x $8^{2}$ x 6)

Evaluate using a calculator

V(cone) = 402.1239

To the nearest tenth, V(cone) = 402.1

Total volume = V(cylinder) + V(cone)

= 804.2 + 402.1

= 1206.3 $ft^{3}$

7 0

3 years ago

Solve for x using the Quadratic Formula: x2 + 4x + 4 = 0 -b + Vb2 - 4ac x= 2a

Irina18 [472]

Answer:

x = -2.

Step-by-step explanation:

In this case, a is 1, b is 4 and c is 4.

-4 plus or minus the square root of 4^2 - 4 * 1 * 4 divided by 2 * 1 = -4 plus or minus the square root of 16 - 16 / 2 = -4 plus or minus the square root of 0 / 2 = -4 / 2 = -2.

x = -2.

Hope this helps!

8 0

3 years ago

20 is what percent of 32 Write as an equation

AlladinOne [14]

20% of 32= 6.4

20% x 32

20 /100 x 32
Reduce the fraction
1 /5× 32
=32/5
=6.4

8 0

3 years ago

Find x, y and ab and how to check if the answer is right

iris [78.8K]

Answer:

x = 2, y = -2, AB = 142

Step-by-step explanation:

The fact that X is the midpoint gives you two relations:

AX = XB

AX + XB = AB

Since you have two unknowns, this number of equations is sufficient to find their values. Substituting the given expressions in the above equations, you have ...

8(3x+5) -3(y-7) -22x = 5x +y +23 -4(x-12)
8(3x+5) -3(y-7) -22x + 5x +y +23 -4(x-12) = 2x -5y +128

Simplifying the first of these can make simplifying the second one easier.

24x +40 -3y +21 -22x = 5x +y +23 -4x +48

2x -3y +61 = x +y +71 . . . . . . we can use this simplification

x -4y = 10 . . . . . . . . . . . . . . . . subtract x+y+61

Now, we can simplify the second equation to ...

2x -3y +61 +x +y +71 = 2x -5y +128

3x -2y +132 = 2x -5y +128 . . . . . simplify the left side

x +3y = -4 . . . . . . . . . . . . . . . add -2x+5y-132

Then the two equations we need to solve are ...

x -4y = 10
x +3y = -4

Subtracting the second from the first, we get

-7y = 14

y = -2

Substituting into the first of these simplified equations, we get

x -4(-2) = 10 . . . . substitute for y

x +8 = 10 . . . . . . .evaluate

x = 2 . . . . . . . . . .subtract 8

So, the solution is (x, y) = (2, -2).

Now, the values of AX and XB are ...

AX = 2x -3y +61 = 2·2 -3(-2) +61 = 71

XB = x+y+71 = 2 +(-2) +71 = 71 . . . . . . . . matches AX, a good sign

AB = 2x -5y +128 = 2·2 -5(-2) +128 = 142 . . . . = AX+XB, another good sign

The desired values are x = 2, y = -2, AB = 142.

_____

You check the answer by filling the values into the expressions given in the problem statement and seeing if you get consistent results. Here, we used the simplified expressions, rather than the original expressions, so if we did the simplification wrong, we may have the wrong answer. It is always best to use the original equations. (A machine solver working with the original equations confirms our result, so "confidence is high.")

4 0

3 years ago