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NikAS [45]
3 years ago
12

You can use a root to undo...

Mathematics
1 answer:
lutik1710 [3]3 years ago
8 0

Answer:

multiplucation problem

Step-by-step explanation:.

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Write your expression for 25 less than 53 and 5
Lilit [14]

(53 +5) - 25 = 33

In math terms, "less than" means to subtract from something. "And" means to add something.

53 and 5 make 58

25 less than 58 is 33

4 0
3 years ago
Help me ASAP for this question
Dafna11 [192]

Answer:

It's the second one.

Step-by-step explanation:

The points on the graph align with the coordinates on the table.

For example, the fist point is (0,2) and the table shows that and the second point lands on (4,3) which is also shown in the table.

6 0
3 years ago
Riding a proportion of 15 is 1 and a half percent of what number​
skad [1K]

Answer:

1000

Explanation:

15=1.5%

1000=100%

10=1%

not really much of an explanation, but i hope you get what i mean by that

7 0
3 years ago
If it takes 2 hours to travel a distance of 120 km, how long would it take to travel 210km if the rate remained the same?
SVETLANKA909090 [29]

Answer:

about 3.5 hrs

Step-by-step explanation:

4 0
3 years ago
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
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