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3 years ago

Can somebody help me out my grade are bad

Mathematics

2 answers:

vesna_86 [32]3 years ago

7 0

Y=3

Since X is equal to 5, we can substitute and do 5+Y=8 then we subtract 5 from both sides and are left with 3

Valentin [98]3 years ago

5 0

Answer:

use ph0tomath it'll answer it for you and show how to do it

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If an equation has an identity how many equations does it have

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3 years ago

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Step-by-step explanation:

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3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (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

3 years ago

A homeowner measured her rectangular backyard as having a length of 40 yards and a width of 25 yards. Her measurements were accu

kotegsom [21]

Answer:

Step-by-step explanation:

7 0

3 years ago

Drex's and Max's ages add up to 29.Seven years ago Drex was twice as old as Max. Find their present ages

nexus9112 [7]

Answer:

There present ages are

Drex age = 17 years

Max age = 12 years

Step-by-step explanation:

Given as ,

The sum of age of Drex and Max = 29 years

And seven years ago Drex was twice as old as max

So , Let the Age of Drex = D and Max age = M

I.e (D - 7 ) = 2 (M - 7)

Or, D - 7 = 2M - 14

Or, 2M - D = 7

And Also D + M = 29

So , from both equations

(2M - D ) + (D + M) = 29 +7

Or, 3M = 36

I.e M = $\frac{36}{3}$ = 12 years

∴ The Age of Drex = 29 - M

The Age of Drex = 29 - 12 = 17 years

Hence there present ages are

The Age of Drex = 17 years

The Age of Max = 12 years Answer

6 0

3 years ago