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Gennadij [26K]
3 years ago
11

Write an expression of 5 Less than a plus k

Mathematics
2 answers:
Zigmanuir [339]3 years ago
7 0

Answer:

5 < a+k

Step-by-step explanation:

This should be it as you said write it out as an expression.

Aleksandr-060686 [28]3 years ago
4 0

Answer:

(a+k)-5 or a-5+k

Step-by-step explanation:

there is not quite enough info, so depending on how you read the question, it could be (a-5)+k; (five less than a) plus k. Or it could be (a+k)-5; five less than(a+k)

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1 4/9 + 2 6/9 is equal to 4 1/9 <br>True or False?​
Law Incorporation [45]

Answer:

True

Step-by-step explanation:

4/9+6/9= 10/9=1 1/9

2+1=3

3+1 1/9=4 1/9

7 0
3 years ago
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You have returned from a trip to Europe and still have $100 euro. How much in U.S. dollars should you receive if the exchange ra
Zina [86]
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8 0
3 years ago
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Please help me please
Murljashka [212]

Answer:

its no big deal

Step-by-step explanation:

no big deal.

5 0
3 years ago
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.

8 0
3 years ago
Simplify the expression.<br> -2 + 2/5b - 1/4b +6=
OleMash [197]

Answer:

3b/20+4

Step-by-step explanation:

-2 + 2/5b - 1/4b + 6

Combine 2/5b and 1/4b to get 3/20b

-2 + 3/20b + 6

Add −2 and 6 to get 4.

4+3/20b

8 0
3 years ago
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