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Nitella [24]
2 years ago
8

I need help to figure this out.

Mathematics
2 answers:
ladessa [460]2 years ago
8 0
1/4 x 5 = 5/20
2/5 x 4 = 8/20
So 1/4 <2/5
2/5 is greater than 1/4
Sloan [31]2 years ago
7 0

Answer:

\frac{1}{4}  =  \frac{5}{20}

\frac{2}{5}  =  \frac{8}{20}

\frac{1}{4}  <  \frac{2}{5}

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-10.7474..... as a fraction
kondaur [170]

In fraction, it would be:

-107474/10000

6 0
2 years ago
What is the value of the 4 in 475
vladimir1956 [14]
The value of 4 in 475 is 4 hundred
8 0
3 years ago
Generally those who are less active have no opportunities to exercise
Alex777 [14]
What is this question?
8 0
2 years ago
Read 2 more answers
What is the length of ST?
adell [148]
Hi there!

Since ST is a tangent to the circle, we can use the relationship: tangent squared = external secant segment x entire secant segment.

WORK:
(I will be using x in place of ST)
x^2 = 7(23 + 7)
x^2 = 7(30)
x^2 = 210
x = squareroot(210) or approximately 14.5 inches

Hope this helps!! :)
4 0
2 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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