What set of reflections would carry trapezoid ABCD onto itself? (1 point) PLEASE HELP

There are 9 gallons of paint for the students to use to paint the fence outside of the schoolyard. Each student has a bucket tha

Vera_Pavlovna [14]

Answer:

27 buckets.

Step-by-step explanation:

If we multiply the 9 gallons of paint by the denominator of the number of gallons that fit into one of the buckets we can discover the answer to the question.

Brainliest please.

6 0

3 years ago

Find the value of x and y

algol13

What do we know about these angles? Immediately, you might notice that (4y-8)° and (16x-4)° share a line. The same is true of (16x-4)° and (14x+4)°. Any straight line forms what's called a straight angle, which measures 180°, so we know that, since they add up to form a straight angle, (14x+4)° and (16x-4)° must add up to 180°. We can use that fact to set up an equation to solve for x:

(14x+4)+(16x-4)=180

After you solve for x, you should look to solve for y. How can we figure out what y is? If you're familiar with the vertical angle theorem, you'll know that all vertical angles (angles that are directly across from each other diagonally) are equal. So we know that 14x+4=4y-8. You can use the value of x you solved for before to solve this one fairly easily, and then you'll have both values.

8 0

3 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

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

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

3 years ago

Use the graph of the derivative of f to locate the critical points x0 at which f has neither a local maximum nor a local minimum

jok3333 [9.3K]

Critical points are where the derivative is 0, i.e. where it crosses the x - axis

The Critical points lies where the derivative is 0, while it crosses the x-axis, SO, in this case the choice 3 looks like best answer for this.

6 0

3 years ago

Write the ordered pair that represents YZ. Then find the magnitude of YZ. y(-4,12), Z(1,19).

ki77a [65]

The magnitude of YZ is 8.6

Explanation:

Y( -4, 12)

Z ( 1, 19)

Magnitude of YZ = ?

We know:

$YZ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$