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Vika [28.1K]
3 years ago
14

7+ 2x divided by 5= 13

Mathematics
2 answers:
Troyanec [42]3 years ago
4 0

Answer:

x=29

Step-by-step explanation:

Given equation :

\frac{(7+2x)}{5} =13

Making the terms of 'x' to one side and constants to the other.

                   (7+2x)=13*5

                       7+2x=65

                            2x=65-7

                              2x=58

                                x=\frac{58}{2}

                                  x=29

So, the value of 'x' is 29.

Masja [62]3 years ago
3 0

Answer:

x = 29

Explanation:

7 + 2x / 5 = 13

Multiply both sides by 5

7 + 2x / 5 = 13 · (5) = (13) · (5)

7 + 2x = 65

Simplify both sides of the equation

2x + 7 = 65

Subtract 7 from both sides

2x + 7 − 7 = 65 − 7

2x = 58

Divide both sides by 2

2x / 2 = 58 / 2

x = 29

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

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Y’all help 3 giving brainlest if you help and give steps (send picture of your work if you can)
cestrela7 [59]

Answer 200 to 520 and 300 to 580.

Step-by-step explanation:

Each one increases by 60.

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Mamont248 [21]
Question 4 is Choice A
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