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3 years ago

the equation of the quadratic formula is represented by the graph y=a(x-3)2-1. what is the value of a?

Mathematics

2 answers:

kramer3 years ago

8 0

Answer:

Step-by-step explanation:

Nana76 [90]3 years ago

6 0

Answer:

The value of a is 2

Step-by-step explanation:

The parabola cuts the graph at (2,1) and is one of the solution of the quadratic equation.

Point (2,1) satisfies the equation.

We can find the value of a by putting the value of x and y

$1=a(2-3)^2-1$

Therefore, the value of a is 2

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single fancy way of saying "Only one man can win.".

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3 years ago

Use spherical coordinates. evaluate (9 − x2 − y2) dv, where h is the solid hemisphere x2 + y2 + z2 ≤ 4, z ≥ 0.

avanturin [10]

In spherical coordinates, we have

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}$

which gives volume element

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

and so the triple integral is given by

$\displaystyle\iiint_H(9-x^2-y^2)\,\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9-\rho^2\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$
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$=\dfrac{592\pi}{15}$

4 0

3 years ago

Use the Polygon tool to draw the image of the given quadrilateral under a dilation with a scale

Nataly [62]

Answer: See the picture attached.

Step-by-step explanation:

A Dilation is defined as a transformation in which the image and the pre-image have the same shape, but their sizes are different.

When the scale factor is greater than 1, the image obtained after the dilation is greater than the pre-image and it is an "Enlargement".

When the scale factor is is between 0 and 1, the image obtained after the dilation is smaller than the pre-image and it is an "Reduction".

In this case we know that the scale factor is:

$scale\ factor=\frac{2}{3}$

Since:

$0$

It is a Reduction.

You can identify that the vertices of the quadritaleral are:

$(-3,0),(-9,-3),(-6,-9),(3,-6)$

So you need to multiply the coordinates of eac vertex by $\frac{2}{3}$ in order to get the coordinates of the image.

Then, you get:

$(\frac{(-3)(2)}{3},0*\frac{2}{3})=(-2,0)\\\\(\frac{(-9)(2)}{3}),\frac{(-3)(2)}{3})=(-6,-2)\\\\(\frac{(-6)(2)}{3},\frac{(-9)(2)}{3})=(-4,-6)\\\\(3*\frac{2}{3},\frac{(-6)(2)}{3})=(2,-4)$