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There exist infinitely many common fractions $\frac{a}{b}$ , where $a > 0$ and $b > 0$ and for which $\frac{3}{5} < \fr

vredina [299]

Answer:

3/5 has the smallest denominator

Step-by-step explanation:

Question:

There exist infinitely many common fractions a/b , where a > 0 and b > 0 and for which 3/5 < a/b< 2/3. Of these common fractions, which has the smallest denominator? Express your answer as a common fraction.

Solution

A Common fraction is a rational number written in the form: a/b. Where a and b are both integers.

The denominator and numerator in this case are greater than zero. That is, they are non zeros.

The least common denominator (LCD) of two non- zero denominators is the smallest whole number that is divisible by each of the denominators.

To find the smallest denominator between 3/5 and 2/3, we would convert the fractions to equivalent fractions with a common denominator by finding their LCM (lowest common multiple).

When comparing two fractions with like denominators, the larger fraction is the one with the greater numerator and the smaller fraction is one with the smaller numerator.

In our solution after comparing, the smaller fraction would have the smallest denominator.

Find attached the solution.

3 0

3 years ago

Find the value of x. Round to the nearest tenth.

mel-nik [20]

Answer:

the answer is 38.5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What are you finding when you are asked what is 60% of 175? the part, the whole or the precent

butalik [34]

Answer:

105 is the part.

Step-by-step explanation:

175 is the whole

60% is the percent

x is the part

175 * 0.60 = 105

3 0

2 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

3 years ago

how much interest does $1,000 earn at an interest rate of 6% compounded semi annually? what is the balance after 2 years

Contact [7]

Answer:

Interest = A - P = 1015 - 1000 = $15

Step-by-step explanation:

3 0

2 years ago