hectictar

If you take a point and make its x-coordinate negative, that will flip the point over the y-axis.

And if you take a point and add  5  to its y-coordinate, that will shift the point up 5 units.

So the image is the pre-image flipped over the y-axis and shifted up  5  units.

\(3\sin\theta\ =\ \sin\theta-\sqrt3\\~\\ 3\sin\theta-\sin\theta\ =\ -\sqrt3\\~\\ 2\sin\theta\ =\ -\sqrt3\\~\\ \sin\theta\ =\ -\frac{\sqrt3}{2}\)

We can look at a unit circle to see what values of  θ  in the interval  [0, 2π]  will satisfy  \(\sin\theta=-\frac{\sqrt3}{2}\) .

Here is a pretty clear unit circle:

\(2\sin\theta\cos\theta+\cos\theta\ =\ 0\)

                                                     Factor   cos θ   out of both terms.

\((\cos\theta)(2\sin\theta+1)\ =\ 0\)

                                                     Set each factor equal to  0  and solve for  θ .

                                                     And remember that we only want solutions for  θ  in the interval  [0, 2π]

\(\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}\)

So the values for  θ  in the interval  [0, 2π]  that satisfy the equation are:

\(\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}\)_

The total number of coins is  65

x + y  =  65       ←This is equation 1

The total value of the coins is $3.80

the total number of cents  =  380

5x + 10y  =  380       ←This is equation 2

From equation 1 we know...

x + y  =  65

x  =  65 - y

From equation 2 we know...

5x + 10y  =  380

                                         We know  x  =  65 - y  so we can substitute  65 - y  in for  x

5( 65 - y ) + 10y  =  380

                                         Distribute  5  to the terms in parenthesees.

325 - 5y + 10y  =  380

                                         Subtract  325  from both sides.

-5y + 10y  =  380 - 325

                                         Combine like terms.

5y  =  55

                                         Divide both sides by  5

y  =  11

Again from equation  1  we know....

x  =  65 - y

                           And we just determined that  y = 11

x  =  65 - 11

x  =  54

Start by finding the equation of the red line.

The red line passes through the points  (1, -2)  and  (3, 1)

slope of red line  =  \(\frac{\text{rise} }{ \text{run} }\ =\ \frac{1--2}{3-1}\ =\ \frac{3}{2}\)

Using the point  (1, -2)  and the slope  \(\frac32\) ,  the equation of the red line in point-slope form is:

y + 2  =  \(\frac32\)(x - 1)

                                 Multiply both sides of the equation by  2 .

2(y + 2)  =  3(x - 1)

                                 Distribute.

2y + 4  =  3x - 3

                                 Subtract  3x  from both sides and add  3  to both sides.

-3x + 2y + 7  =  0

And the red line is dotted, so it is not included in the inequality. So the liner inequality of the graph is either

-3x + 2y + 7  >  0       or       -3x + 2y + 7  <  0

To determine which one, pick a point which we know should make the inequality true and test that point.

We know that the point  (0, 0)  should make the inequality true.

Is it true that     -3(0) + 2(0) + 7  >  0       ?  Yes, it is true that  7 > 0 .

So the linear inequality of the graph is     -3x + 2y + 7  >  0

Let  a  and  b  be the measures of the two other angles of the triangle.

the sum of the interior angles in a triangle  =  180°

a + b + (a + b)  =  180°

2(a + b)  =  180°

a + b  =  90°

the measure of the smallest exterior angle  =  180° - the measure of the largest interior angle

the measure of the smallest exterior angle  =  180° - ( a + b )

the measure of the smallest exterior angle  =  180° - 90°

the measure of the smallest exterior angle  =  90°

It is leftover from when the asker copy and pasted the question. Other sites use $'s to enclose latex, but this site does not. The dollar sign shouldn't be there.

3)

m∠FGE  =  m∠CGD  because they are vertical angles.

\(\frac{\overline{\text{GF}}}{\overline{\text{GE}}}\ \stackrel{?}{=}\ \frac{\overline{\text{GC}}}{\overline{\text{GD}}} \\~\\ \frac{540}{513}\ \stackrel{?}{=}\ \frac{220}{209}\\~\\ \frac{20}{19}\ \stackrel{?}{=}\ \frac{20}{19}\\~\\ \frac{20}{19}\ =\ \frac{20}{19}\)

And   \(\frac{\overline{\text{GF}}}{\overline{\text{GE}}}\ =\ \frac{\overline{\text{GC}}}{\overline{\text{GD}}}\)

So by SAS similarity,  △GFE  is similar to  △GCD