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3 years ago

What is 600,319,287,914 + 297,997,000,026?

Mathematics

2 answers:

geniusboy [140]3 years ago

6 0

898316287940

hope this helps, back to my hole i go

Alex Ar [27]3 years ago

6 0

Step-by-step explanation:

please mark me as brainlest

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Solve for n. (7^2)^4 = n^8

enyata [817]

7^2x4 = n^8
7^8 = n^8
n = 7

4 0

2 years ago

I need help ASAP. Thank you

Fed [463]

Here is the link to the graph. You can scale it and tweak it.
https://www.desmos.com/calculator/mawkflj9ws

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8 0

3 years ago

F/13 = 6.2 what is the value of f

nekit [7.7K]

Answer:

The value of f is 80.6.

Step-by-step explanation:

We are given the equation and asked to solve for f.

If we are given a fraction with a variable in the numerator, we can multiply both sides of the equation by the denominator to isolate it.

For example, if you look at this equation:

$\displaystyle\frac{a}{26}=17.9$

We can multiply both sides by 26 to get the a by itself so the equation can be solved.

$\displaystyle26\times\frac{a}{26}=17.9\times26\\\\a = 465.4$

Therefore, we can apply this same technique to the equation $\frac{f}{13}=6.2.$

$\displaystyle\frac{f}{13}=6.2\\\\13\times\frac{f}{13}=6.2\times13\\\\f = 80.6\ \text{or} \ \frac{403}{5}$

Therefore, the value of f is 80.6.

5 0

3 years ago

Read 2 more answers

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

1 year ago

What is the product in lowest terms? 8/15 * -3/10

german

Multiply the numerators and denominators together:

-24/150

Divide the numerator and denominator by 6:

-24 / 6 = -4
150 / 6 = 25

-4/25

4 0

3 years ago