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4 years ago

∆ABC rotates 90° clockwise about point P to form ∆A′B′C′.

1 answer:

6 0

Assume P(xp,yp), A(xa,ya), etc.
We know that rotation rule of 90° clockwise about the origin is
R_-90(x,y) -> (y,-x)
For example, rotating A about the origin 90° clockwise is
(xa,ya) -> (ya, -xa)
or for a point at H(5,2), after rotation, H'(2,-5), etc.

To rotate about P, we need to translate the point to the origin, rotate, then translate back. The rule for translation is
T_(dx,dy) (x,y) -> (x+dx, y+dy)

So with the translation set at the coordinates of P, and combining the rotation with the translations, the complete rule is:
T_(xp,yp) R_(-90) T_(-xp,-yp) (x,y)
-> T_(xp,yp) R_(-90) (x-xp, y-yp)
-> T_(xp,yp) (y-yp, -(x-xp))
-> (y-yp+xp, -x+xp+yp)

Example: rotate point A(7,3) about point P(4,2)
=> x=7, y=3, xp=4, yp=2
=> A'(3-2+4, -7+4+2) => A'(5,-1)

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3, 10, 17, 24. what is the 50th term in the sequence

Kamila [148]

Check if you can write an equation relating the term number to the actual value
n1=3
n2=10 = 3+7
n3= 17 = n2+7 = n1+7+7 = n1 +2*7
n4= 24 = n1+3*7

so you will notice a pattern
for the x-th term

n_x =3+(x-1)*7

the 50th term would be n_50 = 3+(50-1) * 7

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3 years ago

Which of the following is not equal to sin⁡(-230°)?

xenn [34]

From trigonometry we know that:

if $sin(\theta)=sin(\alpha)$

then, $\theta=n\pi+(-1)^n\alpha$ (where $n$ is an integer)

This can be rewritten in degrees as:

$\theta=n(180^{\circ})+(-1)^n\alpha$ .............(Equation 1)

Now, in our case, $\alpha=-230^{\circ}$

Therefore, (Equation 1) can be written as:

$\theta=n(180^{\circ})+(-1)^n(-230^{\circ})$ ..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of $\theta$ that match.

For n=2, (Equation 2) gives us: $\theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}$ .

Thus, $sin(230^{\circ})=sin(130^{\circ})$

Now, we know that: $-sin(-50^{\circ})=sin(50^{\circ})$

Let n=-1, then:

$\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}$

Thus, $sin(-230^{\circ})=-sin(-50^{\circ})$

Likewise, $sin(-230^{\circ})=sin(50^{\circ})$

Only the last option $sin(-50^{\circ})$ will never match $sin(-230^{\circ})$ because no integral value of $n$ will ever give $\theta=-50^{\circ}$

Thus the last option is the correct option.

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