Answer:
1/2
Step-by-step explanation:
-2y= 8-X
y= -4+X/2
y= X/2-4
recall that: y= mx+c
in the equation y= x/2+C
m=1/2
where m is your slope
2x + 4 = 12
We're simply just trying to isolate x.
So, we must get x onto it's own side of the equal sign :)
Our first step is to subtract 4 from both sides.
2x + 4 - 4 = 12 - 4
Simplify.
2x = 8
Then, we divide both sides by 2.
2x ÷ 2 = 8 ÷ 2
Simplify.
x = 4
----------
To check your work, simply plug in the value of x into x in the original equation.
In this problem, x = 4, so plug in 4 for x.
2x + 4 = 12
2(4) + 4 = 12
Simplify.
8 + 4 = 12
12 = 12
Therefore, x = 4
~Hope I helped!~
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
64
Step-by-step explanation:
can be found by first finding 
then taking that result and putting it into the function 
.
means we are going to take the expression 
and evaluate it for 
:
So 
.
So we have this so far:
.
was found by replacing 
in 
with .
