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3 years ago

Find the missing factor B that makes the equality true. -35x^6=(5x^2)(B)

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

Answer:

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AYOOO HELPPPPP MATHHHH......? THANKKK UUUUU

Gnom [1K]

9514 1404 393

Answer:

choices A and F are part of Leo's system of equations

Step-by-step explanation:

The formula for the amount in an account compounded annually is ...

A = P(1 +r)^t

For the given amounts and variables, that is ...

y = 500(1.025)^x . . . . matches F

The formula for the amount in an account compounded continuously is ...

A = P·e^(rt)

With given values, this is ...

y = 400e^(0.02x) . . . . matches A

3 0

3 years ago

(a) the number 561 factors as 3 · 11 · 17. first use fermat's little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11),

Vitek1552 [10]

LFT says that for any prime modulus $p$ and any integer $n$ , we have

$n^p\equiv n\pmod p$

From this we immediately know that

$a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}$

Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case $11\times17=187=62\times3+1$ , so

$a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3$

Next, $63=21\times3$ , so

$a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3$

Next, $21=7\times3$ , so

$a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3$

Finally, $7=2\times3+1$ , so

$a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3$

We do the same thing for the remaining two cases:

$3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}$

$3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}$

Now recall the Chinese remainder theorem, which says if $x\equiv a\pmod n$ and $x\equiv b\pmod m$ , with $m,n$ relatively prime, then $x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}$ , where ${m_n}^{-1}$ denotes $m^{-1}\pmod n$ .

For this problem, the CRT is saying that, since $a^{561}\equiv a\pmod3$ and $a^{561}\equiv a\pmod{11}$ , it follows that

$a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}$
$\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}$
$\implies a^{561}\equiv 34a\equiv a\pmod{33}$

And since $a^{561}\equiv a\pmod{17}$ , we also have

$a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}$
$\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}$
$\implies a^{561}\equiv562a\equiv a\pmod{561}$

6 0

4 years ago

What is the perimeter of the rectangle in the coordinate plane?

Klio2033 [76]

Here, Length = 14 - 4 = 10 units
Width = 10 - 5 = 5 units

Now, P = 2(l + w)
P = 2(10 + 5)
P = 2(15)
P = 30 units

In short, Your Answer would be Option C

Hope this helps!

7 0

3 years ago

Read 2 more answers

A pond has 150 fish initially, and the population decreases by 8% each day.

V125BC [204]

Answer: not sure.. i got 18.75 so i dont rly know if 18 daily or 19if 19 fish daily then the last day would have 5 left

Step-by-step explanation:

4 0

3 years ago

Solve -4=12p+10/2 can someone please help??

never [62]

So first you would divide the 10 by 2
-4=12p+5

Then you subtract the five from both sides to try to isolate the variable p
-4=12p+5
-5 -5
You then divide both sides by 12 to solve for p and simplify
-9=12p
(-9/12)=(12p/12)
-3/4=p

3 0

3 years ago