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3 years ago

Will 1/4 x 7/8 be larger or smaller than 1/4? Explain your answer.

2 answers:

5 0

It will be smaller than 1/4.

The reason being that 7/8 is a fraction which is less than 1, and that it can only be greater than 1/4 when the multiplying number, is a number which is greater than 1.

Note: When a positive number is multiplied by a number which is less than 1, it makes the positive number to be smaller and

When a positive number is multiplied by a number which is greater than 1, it makes the positive number to be bigger.

zhuklara [117]3 years ago

4 0

$\frac{1}{4}$ × $\frac{7}{8}$ = $\frac{7}{32}$

$\frac{7}{32} = 0.21875$

$\frac{1}{4} = 0.25$

$\frac{1}{4}$ is greater than $\frac{1}{4}$ × $\frac{7}{8}$

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The height of the tunnel at the center is 54 ft, and the vertical clearance must be 18 ft at a point 8 ft from the center. Find

IrinaVladis [17]

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3 years ago

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3 years ago

What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?

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First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

$\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}$

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒ 1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒ 1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

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2 years ago