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1 year ago

The grid shows Figure Q and its image Figure Q′ after a transformation.

Mathematics

1 answer:

Svetach [21]1 year ago

8 0

The transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

<h3>What is transformation?</h3>

A transformation is a general term for four specific ways to manipulate the shape or position of a point, a line, or a geometric figure.

We have,

Coordinates of a pentagon Q.

(2, 4), (3, 7), (7, 5), (5,4), and (4,2).

Coordinates of pentagon Q'.

(-2, -4), (-3, -7), (-7, -5), (-5,-4), and (-4,-2).

We see that the coordinates of each point of the pentagon Q have changed their position as negative coordinates in pentagon Q'

The coordinates of pentagon Q are in (x, y) which can be assumed to be in the first quadrant of the coordinate plane.

The coordinates of pentagon Q' are in (-x, -y) which is in the 3rd quadrant.

From the first quadrant to 3rd quadrant there is a rotation of 180°.

Thus the transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

Learn more about transformation here:

brainly.com/question/4458799

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A triangle has two sides measuring 16 and 21. The included angle is 116°. What is the length of the side opposite the 116° angle

aleksley [76]

Answer: A. 31.5

Step-by-step explanation:

Using law of cosine because it is side angle side

a=16

b=21

angle = 116

c^2=a^2+b^2-2ab cosC, let C be the angle given

c^2=(21)^2+(16)^2-2(16)(21) cos(116)

c^2=441+256-672 cos(116)

i used a ti-89 and answered the right side and got

(you can find a ti-89 simulator online, but for problems like this make sure it is in degree mode)

991.585

c^2=991.585

take square root of both sides

=31.4894, which if you round it up it would be 31.5, so the answer would be A

therefore:

a=16

b=21

c=31.5

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If veronica reed is paid a gross weekly salary of $1036.00, what is her gross annual salary?

Ber [7]

So, (weekly) she receives 1036.00.
The question is asking to find out the (annual) salary which is a year.
There are about 52 weeks in a year.
To find the amount being payed annually we multiply 1036 by 52
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3 years ago

Please help me! Given (x+a)(x+b) = x^2+x-12, what is the value of a^2+b^2

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(x + 4) (x - 3)
a = 4
b = -3

a^2 + b^2 = ...
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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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