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Bogdan [553]
3 years ago
13

What is the length of this line

Mathematics
2 answers:
GenaCL600 [577]3 years ago
8 0

Answer:

13

Step-by-step explanation:

Hi! Use the height and length of the line to form the line into the hypotenuse of a triangle. The Pythagorean Theorem can be used to find the length of the line.

Side a is 5 squares high

Side b is 12 squares long

Side c (the line) is x squares long

a^2+b^2=c^2

5^2+12^2=x^2

x can equal 13 or -13 (because of square rules) but in this case it's thirteen because length can't be negative

Please mark brainliest and have a great day!

and if you have anymore questions please let me know

Bogdan [553]3 years ago
4 0

Answer:

13

Step-by-step explanation:

use pythagorean theorem

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Answer: 100


Step-by-step explanation:

100*100= 10000

The square root of 10000 is 100.

* Hopefully this helps!

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Which statement is true about the points and planes?
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Whats ,, 5x-2x-7y+8x+y= ???​
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Answer:

11x - 6y

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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
Write the equation for the area of a triangle with a height that is 5 more than twice its base using one variable. Please help m
zysi [14]

Answer:

A=b(b+\frac{5}{2} )

Step-by-step explanation:

hey there!!!

we know already that the area of a triangle is given as

A=\frac{1}{2}b *h

where b= base

and h= heigth

Given that the height is 5 times more than twice it base

let h= 2b+5

Now we substitute h=2b+5 in the area of the triangle

A=\frac{1}{2}b * (2b+5)

A= \frac{2b^2}{2} +\frac{5b}{2}

A= b^2+\frac{5b}{2}\\\\A=b(b+\frac{5}{2} )

7 0
4 years ago
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