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AlladinOne [14]
4 years ago
9

Clara lends half her collection of formal attires to her sister, Susan. Clara then buys four more attires. If she has 12 attires

now, how many attires did Clara had initially?
Write the equation to be used to solve this equation.
Mathematics
1 answer:
nirvana33 [79]4 years ago
5 0

Answer:

i think 4 attires

Step-by-step explanation:

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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.

8 0
3 years ago
Kisiki and Jembe are riding on a circular path.Kisiki completes a round in 24 minutes whereas Jembe completes a round in 36 minu
Lelechka [254]
My answer is 12 because I subtract 36 from 24 and got 12
8 0
3 years ago
Aldon and Jamal raised the same amount of money for the school fundraiser. Aldon
Naddika [18.5K]

Answer: $5

Step-by-step explanation: This can be represented by the equation

40 + 12T = 25 + 15T

We just solve for T.

Subtract 12T from both sides

40 = 25 + 3T Subtract 25

15 = 3T Divide by 3

5 = T

7 0
3 years ago
Does anybody know what to do here?
Snezhnost [94]

Answer:

m=6

Step-by-step explanation:

38+8m+4=90

42+8m=90

8m=90-42

8m=48

m=6

6 0
3 years ago
X + 7 = 9<br> What does X equal?
MakcuM [25]
2 i hope this helps ok bye
3 0
3 years ago
Read 2 more answers
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