Answer:
Step-by-step explanation:
i assume x+y=6(1/2)=\frac{13}{2}

the answer will be 20 times 3 that is 60
(a) Using Newton's Law of Cooling,

, we have

where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get

. Integrate both sides to get

.
Since

, we solve for C:

So we get

. Use T(30) = 150 to solve for k:

So

But choose Positive because T > 75. Temp of turkey can't go under.

(b)

Dogs of the AMS.
Answer:
63.2 = y
Step-by-step explanation:
The perimeter is the sum of all the sides
P = 7.8+ y+37.6 + y
171.8 = 7.8+ y+37.6 + y
Combine like terms
171.8 = 45.4 + 2y
Subtract 45.4 from both sides
171.8-45.4 = 45.4 + 2y -45.4
126.4 = 2y
Divide each side by 2
126.4/2 = 2y/2
63.2 = y
A parallel equation (when graphed) will have the same slope, but a different y-intercept.
As long as you keep y = -
x + b, you can input anything for b to solve this question.
Given:
y = -
x - 5
Equation of a parallel line:
y = -
x + 6, y = -
x + 1,356, y = -
x - 8, etc
Example answer you can use:
y = -
x - 8