Answer:
the answer is #4
Step-by-step explanation:
<h3>first you multiply the 3&4 witch is 12 then u keep the y and u add the 2&5 then u see that it is #4</h3>
SOLUTION:
A box has a square base of side x and height y where x,y > 0.
Its volume is V = x^2y and its surface area is
S = 2x^2 + 4xy.
(a) If V = x^2y = 12, then y = 12=x^2 and S(x) = 2x^2 C 4x (12=x2) = 2x2 + 48x^-1. Solve S'(x) = 4x - 48x-2 = 0 to
obtain x = 12^1/3. Since S(x/) ---> infinite as x ---> 0+ and as x --->infinite, the minimum surface area is S(12^1/3) = 6 (12)^2/3 = 31.45,
when x = 12^1/3 and y = 12^1/3.
(b) If S = 2x2 + 4xy = 20, then y = 5^x-1 - 1/2 x and V (x) = x^2y = 5x - 1/2x^3. Note that x must lie on the closed interval [0, square root of 10]. Solve V' (x) = 5 - 3/2 x^2 for x>0 to obtain x = square root of 30 over 3 . Since V(0) = V (square root 10) = 0 and V(square root 30 over 3) = 10 square root 30 over 9 , the
maximum volume is V (square root 30 over 3) = 10/9 square root 30 = 6.086, when x = square root 30 over 3 and y = square root 30 over 3 .
Answer:
Step-by-step explanation:
Look at the picture.
ΔADC and ΔACB are similar. Therefore the corresponding sides are in proportion:
We have
Substitute:
<em>cross multiply</em>
Answer:
a=24
Step-by-step explanation:
= -4
-a= -4*6
-a= -24
a= 24
Domain:
(
−
∞
,
∞
)
,
{
x
|
x
∈
R
}
(
-
∞
,
∞
)
,
{
x
|
x
∈
ℝ
}
Range:
{
y
|
y
=
7
}
