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Annette [7]
3 years ago
9

How l can write a 3-digit number using digits 2,9,4.

Mathematics
2 answers:
ikadub [295]3 years ago
8 0
You can do 942 or 249
Lelechka [254]3 years ago
7 0
492 or 924 or 249 or 942
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I would like to know what this means exactly, feel free to answer it too
LiRa [457]

Answer:

A(-5,2)

Step-by-step explanation:

What they want from you is to basically just flip the graph 90 degrees

4 0
3 years ago
The scores on an exam are listed. If a teacher wants to explain how poorly the class did on the test, would she use the mean, me
Natasha_Volkova [10]
Mean because the mean is an average. That gives an overall understanding of what the entire class did.
4 0
3 years ago
Read 2 more answers
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
PLEASE HELP ASAP GEOMETRY 20 PTS REAL ANSWERS ONLY
kodGreya [7K]

Answer:

Step-by-step explanation:

Use the Law of Cosines to find the measure of angle A from the lengths of the sides.

A = arccos[(b²+c²-a²)/(2bc)] ≅ 29.9°

B = arccos[(a²+c²-b²)/(2ac)] ≅ 54.8°

C = 180 - A - B = 95.3°

7 0
2 years ago
What factors multiply to -80 and add to 6
forsale [732]

80 = -1(80), -2(40), -4(20), -5(4),

Total answers include: -81, 79, -42, 16, -9, and 1.

None of them add up to 6!

Are you doing polynomial equations?

If so, you can solve it the hard way.

x^2 + 6x -80 = 0

(x^2 + 6x + 9) - 80 = 9

(x+3)^2 = 89

(x+3) = 9.43

x = 6.43

If you're not doing those then I'm afraid your question has no answer.

Hope that helps!



7 0
3 years ago
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