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a_sh-v [17]
3 years ago
11

The line with equation y + 2x = 0 coincides with the terminal side of an angle θ in standard position in Quadrant IV .

Mathematics
1 answer:
Ket [755]3 years ago
7 0

Answer:

-2

Step-by-step explanation:


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13.
Nadusha1986 [10]

Answer:

6 vans and 6 buses

Step-by-step explanation:

v+b=12 this equation represents the amount of vehicles

6v + 55b = 366 this equation represents the amount of students per vehicle.

after solving you get v=6 and b=6

5 0
3 years ago
When G(2,-6) is reflected in the line y = -2, the image is located at G |
Mkey [24]

Answer:

The point (2, -6) is four below the line y= -2

The reflection will be four below the line y=-2.

Since y=-2 is a horizontal line, the reflection will be vertical,

so (2,-6) will reflect down 6 spaces vertically. -6 + (4*2) = (-6 + 8) =  2  therefore when we establish 4 below the line we just apply * 2 as a double as its across the line each side of the line to find that below the line = 4 = 4*2 and add it to -6

 

(2,-6) reflected over y=-2 reflects to (2, 2 )

Step-by-step explanation:

4 0
3 years ago
Can somebody help me on this plz
rodikova [14]
The value will be  A= 87
6 0
3 years ago
Read 2 more answers
How can I get help with 2 step equations
Ugo [173]
It takes two steps to solve an equation or inequality that has more than one operation: Simplify using the inverse of addition or subtraction. Simplify further by using the inverse of multiplication or division.
6 0
4 years ago
Write the equation in siope-intercept form for the line that passes through the given pointand is perpendicular to the given equ
yulyashka [42]

For a line in the form:

\begin{gathered} ax+by=c \\ The\text{ slope m is:} \\ m=-\frac{a}{b} \end{gathered}

In this case, for the line 2x + 10y = 20 with a=2 and b=10 the slope is:

\begin{gathered} m_1=-\frac{a}{b}=-\frac{2}{10} \\ m_1=-\frac{1}{5} \end{gathered}

Now, two lines are perpendiculars if the slopes satisfy the following equation:

m_2=-\frac{1}{m_1}

So, for the line we want the slope is:

\begin{gathered} m_1=-\frac{1}{5} \\ m_2=-\frac{1}{m_1}=-\frac{1}{(-\frac{1}{5})}=5 \end{gathered}

Finally, the line pass througth the point (2, 3) with slope m=5, so the equation is:

\begin{gathered} P_1=(2,3),m=5 \\ y=mx+b \\ \text{The P1 must satisfy the equation:} \\ 3=5\cdot2+b \\ b=3-10 \\ b=-7 \end{gathered}

The equation of the line is y = 5x - 7

7 0
1 year ago
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