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2 years ago

What is the image of (−3,4) after the transformation ry=x?

Mathematics

1 answer:

seraphim [82]2 years ago

3 0

Well I would say r×±3=4 that is what I would say it is but I don't know if it's right

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this past Sunday, the giants scored 9 less than twice the cowboys. the packers scored 14 points more than the giants. if the tea

Grace [21]

Let the score of cowboys is x
and giants make score 9 which is twice less than the cowboys score so
giants score will be = 2x -9
and packers scored 14 more than giants that is (2x - 9) + 14
now sum of their scores is equal to 81 it means:
x + (2x - 9) + (2x -9) + 14 = 81
x + 2x - 9 + 2x - 9 + 14 = 81
5x = 81 + 4
5x = 85
x = 17
packers scored = (2x - 9) + 14
= 2 (17) -9 + 14
=38 + 5 = 43 points

5 0

2 years ago

Hurry!! Please help!!!

Virty [35]

(-5+25k-8k-20)-5+25k-8k-20

(17k-25)-17k-25

8 0

2 years ago

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I) Solve the equation sin2x + 3cos2x = 0 (for 0 till 360°)

galina1969 [7]

Get rid of cos2x by dividing both the values. So Sin2x/cos2x +3cos2x/cos2x.
Tan2x = 3
2x = -71.5 so x is -35.6
Use the quadrant method and add 360 twho the two values tou get.

7 0

2 years ago

For the angles α and β in the figures, find cos(α + β)?

Blababa [14]

Answer:

$\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})$

Step-by-step explanation:

Let the hypotenuse of the smaller triangle be h units.

Then; from the Pythagoras Theorem.

$h^2=4^2+2^2$

$h^2=16+4$

$h^2=20$

$h=\sqrt{20}$

$h=2\sqrt{5}$

From the smaller triangle;

$\cos (\alpha)=\frac{4}{2\sqrt{5} }=\frac{2}{\sqrt{5} }$ and $\sin(\alpha)=\frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5} }$

From the second triangle, let the other other shorter leg of the second triangle be s units.

Then;

$s^2+4^2=6^2$

$s^2+16=36$

$s^2=36-16$

$s^2=20$

$s=\sqrt{20}$

$s=2\sqrt{5}$

$\cos(\beta)=\frac{2\sqrt{5} }{6}=\frac{\sqrt{5} }{3}$

and

$\sin(\beta)=\frac{4}{6}=\frac{2}{3}$

We now use the double angle property;

$\cos(\alpha +\beta)=\cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta)$

we plug in the values to obtain;

$\cos(\alpha +\beta)=\frac{2}{\sqrt{5} }\times \frac{\sqrt{5} }{3}-\frac{1}{\sqrt{5} }\times \frac{2}{3}$

$\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})$

3 0

2 years ago

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For which function is it true that y →- ∞ as x →∞

Alecsey [184]

This is not a function because we have an A with many B. It is like saying f (x) = 2 or 4 It fails the "Vertical Line Test" and so is not a function. But is still a valid relationship, so don't get angry with it. It CAN (possibly) have a B with many A. For example sine, cosine, etc are like that. Perfectly valid functions.

4 0

2 years ago

Read 2 more answers