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3 years ago

What is the equation (simplified) the inverse of y=2x^2?

1 answer:

3 0

$2x^2=y\ \ \ |divide\ both\ sides\ by\ 2\\\\x^2=\dfrac{y}{2}\\\\\boxed{x=\sqrt{\frac{y}{2}}}\\\\x=\dfrac{\sqrt{y}}{\sqrt2}=\dfrac{\sqrt{y}\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\to\boxed{x=\dfrac{\sqrt{2y}}{2}}\\\\\\\boxed{f(x)=2x^2\to f^{-1}(x)=\dfrac{\sqrt{2y}}{2}}$

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raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

3 years ago

((Image Included))

Bas_tet [7]

Answer:

=204.13 inches.

Step-by-step explanation:

Using the side x, we can use sine to find the hypotenuse of the triangle with the angle marked 30°.

Sin 30 =x/hypotenuse.

sin 30 = 27/hyp

hyp= 27/sin 30

=54 inches

We can also find the adjacent as follows.

Cos 30 = adjacent/ 54

Adjacent= 54 cos 30

=46.77 inches

Using the angles marked 45 we can find the hypotenuse of the isosceles triangle.

sin 45= x/hypotenuse

sin 45 =27/hypotenuse

hypotenuse = 27/sin 45

=38.18 inches

The hypotenuse of both the triangles making the isosceles triangle are 38.18 inches long.

Perimeter = 54+ 46.77+27+ 38.18+38.18

=204.13 inches.

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2 years ago

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Each side of a square is lengthened by 1 inch. The area of this new, larger square is 9 square inches. Find the length of a side

Savatey [412]

Answer:

91 EASY

Step-by-step explanation:

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3 years ago

What's the scale factor from AABC to ADEF? A) 4/3 B) 3/4 C) 2/5 D) 5/2

guajiro [1.7K]

Answer:

D) 5/2

Step-by-step explanation:

Scale factor = AB/DE = 35/14 = 5/2

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2 years ago

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3 years ago

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