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krok68 [10]
2 years ago
8

How many pairs of two digit natural numbers have a difference equal to 50

Mathematics
1 answer:
dangina [55]2 years ago
3 0

The number exists 99, then 99 - 49 = 50

Therefore, the pair of two-digit numbers exist between 99 and 49.

<h3>What is the difference?</h3>

In math, the word difference exists as the outcome of subtracting one number from another.

Let the first pair of two-digit numbers be xy and the second pair of numbers be yx.

we want a two-digit number, a two-digit number begins from 10 and ends with a two-digit number that exists 99.

Let the difference of 50 exists possible when we take the biggest two-digit number

Therefore, the number exists 99

99 - 49 = 50

Therefore, the pair of two-digit number exists 99 and 49.

To learn more about differences refer to:

brainly.com/question/14335054

#SPJ9

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alexira [117]

Answer:

None of the options are correct, the correct answer is 1,050 boys and 350 girls.

Step-by-step explanation:

A ratio of 3:4 can be turned into a percentage by dividing 3 by 4. This would give us 0.75 or 75%. This means that the remaining 25% (100-75 = 25) are girls in this ratio. If there are 12 boys we can find the total amount by dividing 12 by 0.75 and then multiplying that answer by 0.25 (25%) in order to find the amount of girls in that ratio, like so...

12 * 0.75 = 16 total

16 * 0.25 = 4 girls

If there are a total of 1,400 students we can find how many are boys and how many are girls by simply multiplying by the percentages.

1,400 * 0.75 = 1,050 boys

1,400 * 0.25 = 350 girls.

Apparently, none of the options are correct, the correct answer is 1,050 boys and 350 girls.

6 0
3 years ago
Math question 3, thanks if you help:)
Zanzabum

Answer:

Step-by-step explanation:

Plug in x = 2

g(2) = 3(2)² - 2*2 + 5

         = 3* 4 - 4 + 5

        = 12 - 4 + 5

        = 8 + 5

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3 0
3 years ago
If we approximate the function y=sin(x) with a0+a1 x a2 x^2 +a3 x^3, what is a0,a2,a2,a3?
Rashid [163]

The coefficients a_0,a_1,a_2,a_3 could be chosen to be the coefficients in the Maclaurin series of \sin(x).

We have

y = \sin(x) \approx a_0 + a_1 x + a_2 x^2 + a_3 x^3 \\\\ \implies y(0) = 0 = a_0

y' = \cos(x) \approx a_1 + 2a_2 x + 3a_3 x^2 \\\\ \implies y'(0) = 1 = a_1

y'' = -\sin(x) \approx 2a_2 + 6a_3 x \\\\ \implies y''(0) = 0 = 2a_2

y''' = -\cos(x) \approx 6a_3 \\\\ \implies y'''(0) = -1 = 6a_3

It follows that a_0=0, a_1=1, a_2=0, and a_3 = -\frac16.

7 0
2 years ago
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zaharov [31]
A is the correct answer because tangents do not have any chords
6 0
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Which scatterplot shows the weakest positive linear association?​
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Answer:

Sorry for the late response. The answer is A.

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