Step-by-step explanation:
∆PQR similar to ∆ JKL
JK/KL= PQ/QR
6/12=3/QR
QR=3×2= 6
therefore, tan(R) = PQ/QR = 3/6=1/2
and, tan(J) = KL/JK =12/6=2
tan(J)=4.tan(R)
option D
What is an equation for the linear function whose graph contains the points (2, 5) and (3, 0)?
1 answer:
Answer:
y = -5x + 15
Step-by-step explanation:
We'll look for a linear function of the form y = mx + b, where m is the slope and b is the y-intercept.
We can determine a slope from the two given points by calculating a Rise/Run. Starting at (2,5) and going to (3,0):
Rise = (0 - 5) = -5
Run = (3-2) = 1
Rise/Run = Slope, m) = -(5/1) or -5
Now we have y = -5x + b
To find b, enter either of the two points and solve for b. I'll use (3,0), because I know my zero multiplication and addition tables by heart.
y = -5x + b
0 = -5(3) + b
b = 15
The equation becomes <u>y = -5x + 15</u>
See attachment.
You might be interested in