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3 years ago

7/10÷2 5/8 Do i have to simplify

Mathematics

1 answer:

miv72 [106K]3 years ago

6 0

let's firstly convert the mixed fraction to improper fraction and then divide.

$\bf \stackrel{mixed}{2\frac{5}{8}}\implies \cfrac{2\cdot 8+5}{8}\implies \stackrel{improper}{\cfrac{21}{8}} \\\\[-0.35em] ~\dotfill$

$\bf \cfrac{7}{10}\div \cfrac{21}{8}\implies \cfrac{~~\begin{matrix} 7 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{\underset{5}{~~\begin{matrix} 10\\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{\stackrel{4}{~~\begin{matrix} 8 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{\underset{3}{~~\begin{matrix} 21 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \cfrac{4}{15}$

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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

How to make a mixed or whole number 7/3

eimsori [14]

Answer:

mixed number: 7/3= 2 1/3

4 0

3 years ago

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VMariaS [17]

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3 years ago

#1) Line j contains points (-3,5) and (6,-1). If line p is PERPENDICULAR to linej,

tia_tia [17]

First find slope of line j
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5 0

3 years ago

A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standard deviation of $

Nitella [24]

Answer:

( $74.623, $83.777)

The 90% confidence interval is = ( $74.623, $83.777)

Critical value at 90% confidence = 1.645

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $79.20

Standard deviation r = $10.41

Number of samples n = 14

Confidence interval = 90%

Using the z table;

The critical value that should be used in constructing the confidence interval.

z(α=0.05) = 1.645

Critical value at 90% confidence z = 1.645

Substituting the values we have;

$79.20+/-1.645($10.42/√14)

$79.20+/-1.645($2.782189528308)

$79.20+/-$4.576701774067

$79.20+/-$4.577

( $74.623, $83.777)

The 90% confidence interval is = ( $74.623, $83.777)

5 0

3 years ago