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3 years ago

I NEED HELP ASAP!!!! This is the equation: 7(h−3)+4h Please help!

Mathematics

2 answers:

Karolina [17]3 years ago

7 0

The answer is 11h - 21

timofeeve [1]3 years ago

5 0

Answer:

11h-21

Step-by-step explanation:

Hope this helps :-)

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Help help help please

AURORKA [14]

Answer:

I did this assignment earlier the answer is All Real Numbers.

8 0

3 years ago

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

2 years ago

Find the area of the shape shown below. i need help nowww!!!

vladimir2022 [97]

Lentgh * width = area so, 3.5*9 right?

4 0

3 years ago

Sonic rolls up a slope at 9.4 m/s. after 3.0 he is rolling back down ay 7.4 m/s. how far up the hill is he at this time?

vovikov84 [41]

U= 9.4m/s
v= -7.4m/s (Negative sign because it is in the opposite direction as he is rolling back)
t= ?
s= ?
a= ?
Now, a= v-u/t
= -7.4-9.4÷3
=-5.6
By the second equation of motion.
s= ut+1÷2at*2 ( *2 is the power)
s= 9.4×3+1÷2×-5.6×3*2
= 28.2 +(-25.2)
=3
Therefore s or the distance travelled is 3m.

6 0

3 years ago

Use the substitution method to solve the system of equations

PilotLPTM [1.2K]

You solve the substitution method to solve a system of equality by expressing one variable in terms of the other using one equation, and then plugging this expression in the other(s).

In this case, the first equation gives us a way to express n in terms of m. So, we can replace every occurrence of n in the second equation with the given formula.

The result is

$14m+2n=-8 \iff 14m+2(-7m-4)=-8 \iff 14m-14m-8=-8 \iff -8=-8$

So, the second equation turned to be an equality, i.e. an equation where both sides are the same.

This implies that the system has infinitely many solutions, because every couple $(n,m)$ such that $n=-7m-4$ is a solution to the system, because it satisfies both equations: the first is trivially satisfied, whereas the second is an identity, and as such is satisfied by any value of the variable.

3 0

3 years ago