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Snezhnost [94]
2 years ago
8

Write an algebraic expression for the word phase.

Mathematics
1 answer:
Reika [66]2 years ago
7 0
The answer is 10gh.

The product of g and h means the multiplication of g and h, so it would just be gh.

10 times the product of g and h would mean 10 multiplied by gh, so it would just be gh.
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What is the prime number of 24
tangare [24]
2*2*2*3 is the prime numbers of 24
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3 years ago
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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
 PLSSS HELP IMMEDIATELY!!!!! i’ll give brainiest, i’m not giving brainiest if u leave a link tho. (pls check whole picture!!) a
emmainna [20.7K]

Answer:

(0, 1)

Step-by-step explanation:

All other choices do not match the two possible vertices of the square.

4 0
3 years ago
Find the missing number to make these fractions equal. 6/? = 4/12
Ugo [173]
Answer is 18

Step by step explanation

6/? = 4/12
?=6*12/4=18
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3 years ago
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Please help with math
Sergio [31]

Answer: x = 2

Step-by-step explanation:

x+y=8\\y=3x

Replace y in the first equation by the value of y in the second equation.

x+y=8\\x+3x=8\\4x=8\\x=\frac{8}{4}\\ x=2

4 0
3 years ago
Read 2 more answers
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