Answer with explanation:
We have to prove that, For any integer m, 2 m×(3 m + 2) is divisible by 4.
We will prove this result with the help of Mathematical Induction.
⇒For Positive Integers
For, m=1
L HS=2×1×(3×1+2)
=2×(5)
=10
It is not divisible by 4.
⇒For Negative Integers
For, m= -1
L HS=2×(-1)×[3×(-1)+2]
=-2×(-3+2)
= (-2)× (-1)
=2
It is not divisible by 4.
False Statement.
Hello!
As you can see this line is horizontal. It is 6 units long. Therefore the center is three units in. Therefore the center of this circle is at (1,2).
I hope this helps!
Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
In the first grid more squares are shaded than in the 2nd grid.
