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3 years ago

Please solve 16 and 17! I don't have much time to do my homework tonight.

Mathematics

1 answer:

NeX [460]3 years ago

8 0

16. ∠2 = 53° because it is supplementary.

- You are given the measurement of ∠1, which is 128°. From the picture you can see that ∠1 and ∠2 combined equal 180° (they are supplementary angles).
- So to find ∠2, simply subtract 128° from 180°. The answer is 53°.

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17. 12° because it is vertical.

- Vertical angles, like 48° and 4x in the image, are always equal.
- So if 4x = 48°, then you can divide both sides to get x = 12°.

You might be interested in

Given cosα = −3/5, 180 < α < 270, and sinβ = 12/13, 90 < β < 180

torisob [31]

I got

$- \frac{6 3}{65}$

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

${12}^{2} + {y}^{2} = {13}^{2}$

$144 + {y}^{2} = 169$

$25 = {y}^{2}$

$y = 5$

so our adjacent side is 5.

Cosine is adjacent/hypotenuse so our cos b=5/13.

Using the interval it gave us, Angle a must be in the third quadrant. Since cos is adjacent/hypotenuse and we are given cos a=-3/5. We are going to set up an equation using pythagorean theorem,

$( - 3) {}^{2} + {x}^{2} = {5}^{2}$

$9 + {x}^{2} = 25$

${x}^{2} = 16$

$x = 4$

so our opposite side is 4. sin =Opposite/Hypotenuse so our sin a =4/5.Sin is negative in the third quadrant so

sin a =-4/5.

Now use cosine difference formula

$- \frac{3}{5} \times \frac{5}{13} + - \frac {4}{5} \times \frac{12}{13}$

$- \frac{15}{65} + ( - \frac{48}{65} )$

$- \frac{63}{65}$

Hope this helps

6 0

3 years ago

Help Pleasee<br><br> Reflect shape A in the line x = -2

Cerrena [4.2K]

Answer: Check out the diagram below

The reflected image is shown in red.

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Explanation:

Draw a vertical line through -2 on the x axis. This is the mirror line.

Now focus on the upper right corner of the figure, which is at (-3, -1). Notice how the horizontal distance from this corner point to the mirror line is exactly 1 unit. If we move another 1 unit to the right, then we'll arrive at (-1,-1) which is where the reflected point lands or ends up.

In short, the upper right corner point (-3,-1) reflects over x = -2 to land on (-1,-1)

----------------------

As another example, the upper left corner point (-5, -1) will move exactly 4 spaces to the right to get to the mirror line. Then we move another 4 spaces to the right to get to (2,-1).

So the upper left corner (-5,-1) will ultimately move to (2,-1) after the reflection over x = -2.

Apply these steps to the other corner points and you'll end up with what is shown below.

Take note that a point like A(-5,-1) moves to A'(1,-1), and similar to the other points as well. Also, notice that when going from A to B to C, etc we are moving clockwise. We move counterclockwise when going from A' to B' to C' etc. Reflections always swap the orientation.