Answer:
8x+3y=-6
Step-by-step explanation:
900 is the answer of 2 squared times 15 squared
Answer with explanation:
→→→Function 1
f(x)= - x²+ 8 x -15
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= - 2 x + 8
Put,f'(x)=0
-2 x+ 8=0
2 x=8
Dividing both sides by , 2, we get
x=4
Double differentiating the function
f"(x)= -2, which is negative.
Showing that function attains maximum at ,x=4.
Now,f(4)=-4²+ 8× 4-15
= -16 +32 -15
= -31 +32
=1
→→→Function 2:
f(x) = −x² + 2 x − 3
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= -2 x +2
Put,f'(x)=0
-2 x +2=0
2 x=2
Dividing both sides by , 2, we get
x=1
Double differentiating the function,gives
f"(x)= -2 ,which is negative.
Showing that function attains maximum at ,x=1.
f(1)= -1²+2 ×1 -3
= -1 +2 -3
= -4 +2
= -2
⇒⇒⇒Function 1 has the larger maximum.
Answer:
Let L(n) be the function that gives the amount of logs stacked after n loads.
L(n) = 12 + 8(n-1)
Step-by-step explanation:
Let L(n) be the function that gives the amount of logs stacked after n loads.
Let's call for the moment the first load as L(0)
L(0)= 12
Let r be the number of logs carried in each load, then
L(n) = 12 + nr
Since L(6) (the seventh load) equals 60, we have
60 = 12 + 6r, and r = 8.
So a function for the number of loads starting from n=0 would be
L(n) = 12 + 8n
If we want to start with n=1, we simply change the variable
L(n) = 12 + 8(n-1) (n=1,2,3,...).
So L(1) = 12, L(2) = 20, L(3) = 28,...L(7) = 60 and so on.
