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4 years ago

Which expressions are equivalent to 647 x 39

Mathematics

2 answers:

stellarik [79]4 years ago

8 0

25233 is your answer for this problem

ser-zykov [4K]4 years ago

4 0

Answer:25233

Step-by-step explanation:

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Help me please! <br> I need to type the word but I don’t know what it is

Y_Kistochka [10]

Answer: Exponential Function

Step-by-step explanation: This is a(n) exponential function becuase each input is going at a constant rate of x^2 which is an exponent. Meaning the output is squared by the given number (which is x).

Make me brainliest if it is right.

4 0

3 years ago

The area of a circle is 31.4 cm^2 what’s the radius

34kurt

R=square root of 10 is the answer

7 0

4 years ago

Read 2 more answers

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

4 years ago

Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why th

Veseljchak [2.6K]

If the coefficient matrix has a pivot in each column, it means that it is shaped like this:

$A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]$

So, the correspondant system

$Ax = b$

will look like this:

$\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]$

This turn into the following system of equations:

$\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}$

The last equation is solvable for $x_4$ : we easily have

$x_4=\dfrac{b_4}{a_{4,4}}$

Once the value for $x_4$ is known, we can solve the third equation for $x_3$ :

$x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}$

(recall that $x_4$ is now known)

The pattern should be clear: you can use the last equation to solve for $x_4$ . Once it is known, the third equation involves the only variable $x_3$ . Once

4 0

4 years ago

Steve can complete the 100m dash in 10 seconds while Paul can run it in 12 seconds. How does Steve's time compare to Paul's?

Lana71 [14]

Let

x---------> Steve's time

y--------> Paul's time

we know that

$x=10\ s\\y=12\ s$