Answer:
$y=(7/2)x+20$
Step-by-step explanation:
SInce the gradient of the first line is $-2/7$ then the gradien of the perpendicular line is $7/2$.
Therefore by the point slope formula the line that we are looking for is
$y-6=(7/2)(x+4)$
$y=(7/2)x + 20$
The answer is c. 0.10. Week 4 is plotted at 0.35 and Week 2 is plotted at 0.25. Difference means subtract. 0.35 - 0.25 = 0.10. Hope this helps!
Answer:
10.50°C
Step-by-step explanation:
Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3
The rate of change in temperature of the bug path can be expressed using the composite formula:
dT/dt = Tx(dx/dt) + Ty(dy/dt)
If x = 2+t; dx/dt = 1
If y = 1+12t; dy/dt = 1/2
Substituting the parameters gotten into dT/dt we will have;
dT/dt = 9(1)+3(1/2)
dT/dt = 9+1.5
dT/dt = 10.50°C/s
Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s
Answer:
13/14 is greater
Step-by-step explanation:
Find the lcm for both:
<u>9 x 7 </u> = <u>63</u>
10 x 7 70
<u>13 x 5 </u> = <u>65</u>
14 x 5 70
so therefore 13/14 is greater since it has a greater value when finding the lcm
Answer:
It is <u>C</u> Or the <u>Third Graph</u>
Step-by-step explanation:
correct answer for edge 2020
