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GenaCL600 [577]
3 years ago
13

Order these fractions from leastto greatest.2/30, 2/5, 2/4​

Mathematics
2 answers:
zavuch27 [327]3 years ago
7 0

Answer:

From least to greatest is 2/30, 2/5, 2/4

<em>Hope I Helped</em>

kifflom [539]3 years ago
5 0

Answer:

<h2>2/30 < 2/5 < 2/4</h2>

Step-by-step explanation:

We notice that the three fractions have the same numerator 2

<u><em>remark :</em></u> to compare two fractions with the same numerator all we have to do is comparing their denominators

the greater the denominator is the less the fraction is.

since 30 > 5 > 4 then 2/30 < 2/5 < 2/4

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Y – 3(2y – 7) = 76<br><br> solve for Y!!
Travka [436]

Answer:

y=-11

Step-by-step explanation:

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3 years ago
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What is the product of (x-2)(x-6)? *
belka [17]

Answer:

x^2-4x+12

Step-by-step explanation:

(x-2)(x-6)

x(x-6)-2(x-6)

x^2-6x-2x+12

x^2-4x+12

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3 years ago
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How do you find the volume of the solid generated by revolving the region bounded by the graphs
d1i1m1o1n [39]

Answer:

About the x axis

V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

Step-by-step explanation:

For this case we have the following functions:

y = 2x^2 , y=0, X=2

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on  y from 0 to 8.

We can find the area like this:

A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= 4\pi \int_{0}^2 x^4 dx

V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

For this case we need to find the function in terms of x like this:

x^2 = \frac{y}{2}

x = \pm \sqrt{\frac{y}{2}} but on this case we are just interested on the + part x=\sqrt{\frac{y}{2}} as we can see on the second figure attached.

We can find the area like this:

A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= \pi \int_{0}^2 64-32x^2 +4x^4 dx

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \frac{\pi}{2} \int_{0}^8 y dy

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

6 0
3 years ago
Simplify the expression ​
finlep [7]

Answer:

\frac{2*x - 2}{2*x}  - \frac{3*x + 2}{4*x} = \frac{x - 6}{4*x}

Step-by-step explanation:

We have the expression:

\frac{2*x - 2}{2*x}  - \frac{3*x + 2}{4*x}

The first thing we want to do, is to have the same denominator in both equations, then we need to multiply the first term by (2/2), so the denominator becomes 4*x

We will get:

(\frac{2}{2} )\frac{2*x - 2}{2*x}  - \frac{3*x + 2}{4*x} = \frac{4*x - 4}{4*x}  - \frac{3*x + 2}{4*x}

Now we can directly add the terms to get:

\frac{4*x - 4}{4*x}  - \frac{3*x + 2}{4*x} = \frac{4*x - 4 - 3*x - 2}{4*x}  = \frac{x - 6}{4*x}

We can't simplify this anymore

3 0
3 years ago
The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distr
Tanya [424]

Complete Question

The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Test H_o : p =  0.28 vs H a : p <  0.28when the sample has n 900 , and \^ p = 0.216 with SE=  0.01 . Find the value of the standardized z -test statistic. Round your answer to two decimal places.

Answer:

The value is  z =  -6.40

Step-by-step explanation:

Generally the standardized test statistics is mathematically represented as

     z =  \frac{\^ p  -  p }{ SE }

=>  z =  \frac{0.216  -  0.28 }{ 0.01  }

=>  z =  \frac{0.216  -  0.28 }{ 0.01  }

=>  z =  -6.40

4 0
2 years ago
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