4 x 2 tens = 80, 28 x 14 = 392
Look at the picture.
The point-slope form of a line:

We have the points (0, 1) and (2, 0). Substitute:

Answer: 
Answer:
sorry i dont know
Step-by-step explanation:
but do the ones in the () first always
i. The Lagrangian is

with critical points whenever



- If
, then
. - If
, then
. - Either value of
found above requires that either
or
, so we get the same critical points as in the previous two cases.
We have
,
,
, and
, so
has a minimum value of 9 and a maximum value of 182.25.
ii. The Lagrangian is

with critical points whenever
(because we assume
)



- If
, then
. - If
, then
, and with
we have
.
We have
,
,
, and
. So
has a maximum value of 61 and a minimum value of -60.
Answer:
(x - 4)(x + 4)(x - 1)(x + 1)
Step-by-step explanation:
Given
- 17x² + 16
Use the substitution u = x², then
u² - 17u + 16
Consider the factors of the constant term (+ 16) which sum to give the coefficient of the u- term (- 17)
The factors are - 16 and - 1, then
u² - 17u + 16
= (u - 16)(u - 1) ← replace u by x²
= (x² - 16)(x² - 1) ← both factors are difference of squares
= (x - 4)(x + 4)(x - 1)(x + 1)