Y= 3/4x - 2 (-2,10)= (x1,y1)
The slope is -4/3 of the perpendicular line:
Then:
y-y1=m(x-x1)
y-10=(-4/3)(x+2)
y=-(4/3)x -8/3 +10
y= -(4/3)x +22/3 (That's the answer)
Answer:
left to right
yes no no no yes yes yes yes yes yes yes no
Step-by-step explanation:
the x values can only have 1 y
Y's can have as many X's
Answer:
(12)3=36
Step-by-step explanation:
i don't know.. I think this is what the question is trying to say XD
Assuming simple interest (i.e. no compounding within first year), then
At 6%, interest = 10000*0.06=$600
At 9% interest = 10000*0.09 = $900
Two ways to find the ratio
method A. let x=proportion at 6%
then
600x+900(1-x)=684
Expand and solve
300x=900-684=216
x=216/300=0.72 or 72%
So 10000*0.72=7200 were invested at 6%
10000-7200=2800 were invested at 9%
method B: by proportions
Ratio of investments at 6% and 9%
= 900-684 : 684-600
=216 : 84
= 18 : 7
Amount invested at 6% = 18/(18+7) * 10000 = 0.72*10000 = 7200
Amount invested at 8% = 7/(18+7)*10000=0.28*10000=2800
Graph 1 is related to table C because the first 3 values are increasing, and the 4th value decreases. Or because at 1PM, 3PM, and 5PM, the temperature was increasing, but at 7PM the temperature decreased. Graph 1 shows the first 3 points increasing, and then decreasing at the 4th point.
Graph 2 is related to table A because as the time increases/goes on, the temperature decreases exponentially/continues to decrease at a higher rate than before. From 1-3PM the temperate decreases by 2°F, from 3-5PM it decreased by 8°F, from 5-7PM the temperature decreased by 17°F.
Graph 3 is related to table B because as the time increases/goes on, the temperature decreases at a steady rate of 1°F every 2 hours.
