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bekas [8.4K]
4 years ago
13

How can you solve this problem? Please help!!

Mathematics
1 answer:
polet [3.4K]4 years ago
7 0
A. 13 hours
b. 0.52 hours
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102.8 divided by 4 And how u got it :)
eimsori [14]

Answer:

102.8/4=25.7

I got it because I used a calculator

Step-by-step explanation:

7 0
3 years ago
HELP PLEASE ILL GIVE BRAINLIEST!
yulyashka [42]

It should be 3 because you are going to be using the formation of 3√ 3•3•3

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4 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-%2022x%20%2B121" id="TexFormula1" title="x^{2} - 22x +121" alt="x^{2} - 22x +12
cupoosta [38]

x2 - 22x + 121

= x2 - 11x - 11x +121

= x(x-11) - 11 ( x -11)

= (x-11) (x-11)

6 0
3 years ago
Solve the inequalities <br> x2 &lt; 3 - 2x
ddd [48]

Inequality Form:

−3 < x < 1

3 0
2 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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