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ololo11 [35]
2 years ago
5

Y= 4x + 2 where the domain is the set of all of the positive even numbers less than 10.

Mathematics
1 answer:
inysia [295]2 years ago
5 0

The range of the function y = 4x + 2 where the  domain is the set of all of

the positive even numbers  less than 10 is {10, 18, 26, 34}

The given equation is:

y  =  4x  +  2

The domain is the set of all of  the positive even numbers  less than 10.

That is, x = {2, 4, 6 ,8}

To get the range, find the values of y for x = 2, 4, 6, and 8

For x = 2:

y = 4(2)  +  2

y  =  8  +  2

y   =  10

For x = 4:

y = 4(4)  +  2

y  =  16  +  2

y   =  18

For x = 6:

y = 4(6)  +  2

y  =  24  +  2

y   =  26

For x = 8:

y = 4(8)  +  2

y  =  32  +  2

y   =  34

Therefore, the range = {10, 18, 26, 34}

Learn more here: brainly.com/question/18479953

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2 years ago
We are given a sequence of five non-zero numbers, where the sum of each term and its neighboring terms is 15 or 25. Find the sum
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We have a sequence that meets the given criteria, and with that information, we want to get the sum of all the terms in the sequence.

We will see that the sum tends to infinity.

So we have 5 terms;

A, B, C, D, E.

We know that the sum of each term and its neighboring terms is 15 or 25.

then:

  • A + B + C = 15 or 25
  • B + C + D = 15 or 25
  • C + D + E = 15 or 25

Now, we want to find the sum of all the terms in the sequence (not only the 5 given).

Then let's assume we write the sum of infinite terms as:

a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + ...

Now we group that sum in pairs of 3 consecutive terms, so we get:

(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + ...

So we will have a sum of infinite of these, and each one of these is equal to 15 or 25 (both positive numbers). So when we sum that infinite times (even if we always have the smaller number, 15) the sum will tend to be infinite.

Then we have:

(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) +  ... \to \infty

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brainly.com/question/21885715

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Answer:

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Step-by-step explanation:

1.

(3,1) and\:perpendicular \:to \: y = -x+1\\(3,1) = (x,y)\\m = -1\\In\: perpendicularism ; m_2 = \frac{-1}{m_1} \\m_2 = \frac{-1}{-1} \\m_2 =1\\y = mx+c\\1 = 1(3) +c\\1=3+c\\1-3=c\\-2 =c\\Substitute \:new \:values\:into\\y =mx+c\\y = 1x -2\\y = x-2

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