For this case, the area is given by:
A = x * (200-2x)
Rewriting:
A = 200x-2x ^ 2
Deriving the expression we have:
A '= 200-4x
Equaling zero we have:
200-4x = 0
We clear x:
4x = 200
x = 200/4
x = 50 feet
Then, the maximum area is:
A (50) = 50 * (200-2 * 50)
A (50) = 5000 feet ^ 2
Answer:
the maximum possible area that can be enclosed with his 200ft of fencing is:
A (50) = 5000 feet ^ 2

8 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

How much times can 6 go into 48

DENIUS [597]

6 can go into 48 8 times. How I figured out?
Start multiplying from a small number like
6×6= 36
6×7=42
6×8= 48
until you see your number 48, or a number that is lower that 48, but can not go any higher than 48.

6 0

3 years ago

This question relates to concepts covered in Lectures 1 & 2. You can use any of the excel files posted to work through the q

liubo4ka [24]

Answer:

mean of this demand distribution = 100

Step-by-step explanation:

To find the mean of this demand distribution;

Mean = Expected vale = E[x]

for discrete provability function,

we say E[x] = ∑(x.p(x))

x p(x) x.p(x)

10 0.1 1

30 0.4 12

60 0.4 24

90 0.7 63

∴ ∑(x.p(x)) = ( 1 + 12 + 24 + 63 )

∑(x.p(x)) = 100

7 0

2 years ago