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uranmaximum [27]
3 years ago
14

How do I multiply at 9/25×1/6

Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0

Answer:

3/50

Step-by-step explanation:

It's not as hard as you might think! You can just multiply the denominator and numerators separately. So in this case, you just need to do 9*1 and 25*6. This means that the answer is just 9/150, simplify it and you get, 3/50!

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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
Find what x is? 5×3>4+8+x a good help for your brain! ​
MaRussiya [10]

Answer:

x = 2.25 or any number greater

Step-by-step explanation:

5x3 > 4 + 8 + x

Make them equal to each other to get x

5x3 = 4 + 8 + x

(5x)(3) = 12 + x

- x - 3   - 3  - x

4x = 9

\frac{4x}{4} = \frac{9}{4}

x = 2.25

Once you get x plug it in to get the truth of the expression

6 0
3 years ago
Solve for b, i already solved for A <br> will give brainliest
larisa86 [58]

Answer:

b=23 degrees

Step-by-step explanation:

a was 66 degrees so b=23 degrees   180-90-66=23

6 0
3 years ago
I’ve been trying to figure out this question on my homework and I’ve been confused for days now.
dsp73

Answer:

18m^12n^3

Hope this helps!!

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
20 points show work pls im in a hurry
Gwar [14]

Answer: since it doesnt say 620 is included , we can conclude that x>620, not x>=620 so the 8th week would be cut just short, so answer is 7

aka

fewer than 8 weeks

Step-by-step explanation:

1200-620=580

580/72.50=8

8 0
2 years ago
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