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chubhunter [2.5K]
3 years ago
14

HELP! 15 POINTS AND BRAINLIEST!

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

8x squared 3

Step-by-step explanation:

hope this helps hope you get it right answer is d

Artyom0805 [142]3 years ago
6 0

Answer:

d

Step-by-step explanation:

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Interquartile range is the range is the number in the dead center, you have to divide the number line into 2 sections. The middle of everything and the middle of both section is the interquartile range

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Given the vectors w = &lt;-5, 3&gt; and z = &lt;1, 4&gt;, find the results of the vector subtractions. -w − z =
kvasek [131]

Answer:

< 4, - 7 >

Step-by-step explanation:

- w - z

= - < - 5, 3 > - < 1, 4 > ← multiply each component in w by - 1

= < 5, - 3 > - < 1, 4 > ← subtract corresponding components

= < 5 - 1, - 3 - 4 >

= < 4, - 7 >

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3 years ago
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Lina20 [59]

Answer:

The 3rd one

Step-by-step explanation:

It's a translation. The rest are reflections.

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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.

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3 years ago
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