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3 years ago

How do you work out the 10th term?

Mathematics

2 answers:

WARRIOR [948]3 years ago

6 0

205 is the answer I’m pretty sure ???

poizon [28]3 years ago

3 0

Answer:

I believe that you multiply 5 x 10 because you multiply any of the terms by 5.

Step-by-step explanation:

Like another example is the 41st term which is 5 × 41 = 205

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A B means:

SVETLANKA909090 [29]

Answer:

A is a subset of B i believe that's the answer

8 0

3 years ago

The ratio of Michael and Neil’s ages is 7:6. If Neil is really 42 years old, how old is Michael?

Anit [1.1K]

Answer:

49 years old.

Step-by-step explanation:

By proportion 7/6 = M/42 where M is Michael's age.

6M = 7 * 42

M = (7*42) / 6

M = 7 *7

= 49.

6 0

3 years ago

What is the main idea of this passage from the

ella [17]

Answer: B. The United States might not have been able to stop the attacks, but it could have made a better effort to do so.

Step-by-step explanation:

6 0

3 years ago

Given that a function, g, has a domain of -20 sxs 5 and a range of 5 s 39 s 45 and that g(0) = -2 and g(-9) = 6, select the stat

luda_lava [24]

Answer:

option (B) and (D)

Step-by-step explanation:

It's just basic of domain and range just check out the domain and range you can get your answer .

Only option (B) and (D) satisfy in the given range and domain

8 0

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