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2 years ago

Point e is the midpoint of ca and bd. which statements about the diagram are true?

Mathematics

1 answer:

defon2 years ago

8 0

Answer:

Step-by-step explanation: i don’t know to be honest

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The ratio of boys to girls in a sophomore class of 460 students is 11:12. find the number of boys.

lukranit [14]

11 : 12.....11 + 12 = 23

boys : 11/23 * 460 = 5060/23 = 220 <===
girls : 12/23 * 460 = 5520/23 = 240

7 0

3 years ago

What is 16,107,320 written in word form

Colt1911 [192]

Sixteen million one hundred seven thousand three hundred and twenty

8 0

2 years ago

Solye for x. 8(x + 1) - 3[X + 4) = 7(2 - x)

kramer

Answer:

3/2

Step-by-step explanation:

distribute 8,-3,7

4 0

3 years ago

What is the least common multiple of 8 and 12? 24 48 96

Anna007 [38]

Answer:

The least common multiple of 8 and 12 is 24

Explanation:

Find and list the multiples each number until the first common multiple is found.

This is the lowest common multiple .

Multiples of 8

8, 16, 24, 32,40

Multiples of 12

12,24, 36,48

Therefore,

LCM of 8 and 12 is 24

3 0

3 years ago

Read 2 more answers

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