Please provide the answer for the x and z as seen in the image

Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.

torisob [31]

Fermat's little theorem states that
$a^p$ ≡a mod p

If we divide both sides by a, then
$a^{p-1}$ ≡1 mod p
=>
$a^{17-1}$ ≡1 mod 17
$a^{16}$ ≡1 mod 17

Rewrite
$a^{1000000}$ mod 17 as
$=(a^{16})^{62500}$ mod 17
and apply Fermat's little theorem
$=(1)^{62500}$ mod 17
=>
$=(1)$ mod 17

So we conclude that
$a^{1000000}$ ≡1 mod 17

6 0

3 years ago

What is the value of x ? Circle the correct 2.5x=35. X=14 X=16

scoundrel [369]

Answer:

All you have to do is divide both sides by 2.5 to find your answer.

2.5x / 2.5 = 35 / 2.5

x = 14

The answer is X = 14.

I hope this helps!

-Mikayla

8 0

3 years ago

Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

Step 1: Subtract x²+12 from both sides.

−7x−(x²+12)=x²+12−(x²+12)

−x²−7x−12=0

Step 2: Factor left side of equation.

$(-x-3)(x+4)=0$

Step 3: Set factors equal to 0.

$-x-3=0$ or $x+4=0$

Get your answer:

$x=-3$ or $x=-4$

3 0

3 years ago

12 whites 4 blacks 3 blue 1 red. Find each experimental probability as a fraction? 1.)p(white)

Dafna1 [17]

Experimental probability is if an experiment was actually run. If those values are the result of an experiment, then the probability is desired value (white) over total possibilities.

$\frac{12}{20} = \frac{3}{5}$ or 0.6

5 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago