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3 years ago

If M is a multiple of 3 and 7 then it is a multiple of 21

Mathematics

1 answer:

Jobisdone [24]3 years ago

8 0

Answer:

It will be true , it is a multiple of 21

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How do u order decimals ??

Furkat [3]

Answer:

To order decimals, you first need a range of values.

Step-by-step explanation:

For example, if there are 5 values: 0.25, 0.87,0.19,0.89 and 0.98

First, look at the first digit after the decimal. The smallest digit is the smallest so far. In this case, it is 0.19 as '1' was the smallest digit. Second would be 0.25 because 2 is the second smallest digit here.

Now, we have 3 remaining numbers: 0.87, 0.89 and 0.98

You can notice that both 0.87 and 0.89 have the same first digit after the decimal.

Do fix this problem, look at the second digit and decide the smallest one. In this case, it is 0.87.

This means the order will go like this if it is smallest to largest: 0.19, 0.25, 0.87, 0.89, 0.98.

If it is largest to smallest, do it the other way around.

Hope this helped!

4 0

2 years ago

20. What is the equation of the line that passes through the points (-2, 2) and (0,5)?

ludmilkaskok [199]

B
Slope=(y2-y1)/(x2-x1)
Slope=(5-2)/(0- -2)
Slope=3/2

5 0

3 years ago

In a right triangle with a hypotenuse equal to 5, one leg is one more than the other. Find the lengths of the legs of be triangl

Kryger [21]

Answer:step 1:

zuk my b4lls

step 2: sh6t up

Step-by-step explanation:

7 0

3 years ago

Need answers please

Andrews [41]

Answer:

Step-by-step explanation:

you are adding

6 0

3 years ago

Read 2 more answers

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