All categories

4 years ago

Can someone do 11??? Please help I don’t want to fail my class

Mathematics

1 answer:

jolli1 [7]4 years ago

4 0

<h3>Answer: choice A) </h3><h3>tan(Y) = 6/7 and tan(Z) = 7/6</h3>

Recall that the tangent of an angle is equal to the ratio of the opposite over adjacent sides

tan(angle) = opposite/adjacent

For reference angle Y, the opposite side is XZ = 6 and the adjacent side is XY = 7, so,

tan(Y) = XZ/XY = 6/7

For tan(Z), things will swap. The opposite side is now XY = 7 and the adjacent side is now XZ = 6

tan(Z) = XY/XZ = 7/6

We do not use the hypotenuse sqrt(85) at all in this problem because the tangent ratio does not involve the hypotenuse.

You might be interested in

3. The volume of a book is 60 cubic inches. If a large

saveliy_v [14]

Answer:

720 in^3.

Step-by-step explanation:

12 * 60 = 720.

6 0

2 years ago

Read 2 more answers

What is the slope between the points (8, -2) and (-7, 10)

lara [203]

Slope formula:
(y2-y1)/(x2-x1)

Substitute in the points into the slope formula.

Slope= [10-(-2)] / [(-7)-8]
Slope= (10+2) / (-15)
Slope= 12/-15
Slope= -12/15
Simplify (3 goes into each)
Slope= -4/5

Hope this helps!! :)

8 0

3 years ago

How do you distribute -3(5-r)

fredd [130]

The answer is: -3r+15r
you will multiply -3*5 first then -3*-r

6 0

3 years ago

Sandra had some ribbon to make bows for her hair. she cut the ribbon into 8 pieces. each piece was of a foot long. how long was

DochEvi [55]

8 feet....................

6 0

3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

3 years ago

Read 2 more answers