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elena55 [62]
3 years ago
12

You are designing a rectangular and closure with a rectangular sections separated by parallel walls if you have 2500 feet of fen

cing what is the maximum area that can be enclosed
Mathematics
2 answers:
erica [24]3 years ago
8 0

Answer:

781250

Step-by-step explanation:

Let x = length of fenced side parallel to the side that borders the river

and y = length of each of the other two fenced sides  

Then, x + 2y = 2500

<=>  x = 2500- 2y

Area = xy = y(2500-2y)

            = -y^{2} + 2500y

The graph of the area function is a parabola opening downward.

The maximum area occurs when y = -2500/[2(-2)] = 625

                                             x = 2500-2y = 1250

The maximum area that can be enclosed = xy = 1250*625 = 781250

Gnoma [55]3 years ago
8 0

Step-by-step explanation:

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Label the points, lines, and planes to show AB and line m perpendicular to each other in plane S at point B. plane S
Romashka-Z-Leto [24]

Answer:

Please fins attached the required labelled drawing

Step-by-step explanation:

The given parameters are;

1) Lines AB and m are both coplanar on plane S and are perpendicular to each other intersecting at point B

2) Plane R and plane S intersect on line p

3) Line AB and line p are perpendicular to each other and both intersect with line n at point A

6 0
3 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
4 years ago
2 is more than a number v
Snowcat [4.5K]
2 is more than a number v

 2 > v is your equality

this means that <em>v</em> can be the number 1 or less

hope this helps
8 0
4 years ago
An older person is 4 years older than 7 times the age of a younger person. The sum of their ages is 44. Find their ages
Fynjy0 [20]

Answer:

5 yr and 39 yr

Step-by-step explanation:

Let x = the age of the younger

then 7x + 4 = the age of the older

Now x + 7x + 4 = 44

8x + 4 = 44

8x = 40

x = 5

7x + 4 = 7(5) + 4 =  35 + 4 = 39

Check:  5 + 39 = 44

8 0
3 years ago
Please tell me where I went Wrong. I am Confused
aivan3 [116]
The final answer would be 39/19 it doesn't simplify any more and if it did the 19 would be reduced as well as the 39
6 0
3 years ago
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