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Anton [14]
3 years ago
10

A triangle has vertices A(-4,2), B(-2,3) and C(0,2). Find the vertices of the triangle if it is reflected over the x- axis and t

hen dilated by a scale factor of 2.
Mathematics
1 answer:
VashaNatasha [74]3 years ago
3 0
Reflecting over the x-axis involves multiplying all y-coordinates by -1: A(-4, -2), B(-2, -3), C(0, -2).Dilating by a scale factor of 2 doubles all coordinates: A(-8, -4), B(-4, -6), C(0, -4).

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In ΔNOP, o = 660 inches, p = 510 inches and ∠N=152°. Find the length of n, to the nearest inch.
VikaD [51]

Answer:

1136

Step-by-step explanation:

deltamath

7 0
3 years ago
Which of the binomials below is a factor of this trinomial? -3x^2-21x+24 x-4 x+1 x+4 x-1
OleMash [197]

Answer:

-3x^2-21x+24

Step-by-step explanation:

Kindly separate the answer choices:  (x - 4), (x + 1), (x + 4) , (x - 1).  The trinomial is thus -3x^2-21x+24, which can be factored as follows:  -3(x^2 + 7x - 8)

Then we have -3x^2-21x+24 = -3(x^2 + 7x - 8), which factors further into:

-3(x + 8)(x -1)  and so we conclude that the only binomial that is a factor of

-3x^2-21x+24 is (x - 1).

5 0
3 years ago
The function g(x)=(x-2^2. The function f(x)=g(x) +3
Ksenya-84 [330]
You don't provide the instructions for this problem, leaving it up to me to guess what you might want.

Note that g(x) should be written as g(x) = (x-2)^2, whereas f(x) = g(x) + 3 (as presented).

If we let g(x) be the input to f(x), we get the "composition" of g and f:

f(x) = g(x) + 3 = (x-2)^2 + 3.  You could leave the answer as is or you could expand (x-2)^2:  f(x) = x^2 - 4x + 4 + 3 (and so on).
8 0
4 years ago
(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a
Mazyrski [523]

The length of a curve C given parametrically by (x(t),y(t)) over some domain t\in[a,b] is

\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

In this case,

x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)

y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)

So we have

\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t

and the arc length is

\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt

We have

\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4

where n is any integer; this tells us \sin(4t)\ge0 on the interval \left[0,\frac\pi4\right] and \sin(4t) on \left[\frac\pi4,1\right]. So the arc length is

=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)

=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)

=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}

7 0
3 years ago
This is so confusing
Gala2k [10]

5*9=45

3*9=27

45-27=8

9*2=18

Therfeore the answer is A.

4 0
3 years ago
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