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Serggg [28]
3 years ago
15

What is the product? 2y/y-3 multiply 4y-12/2y+6

Mathematics
1 answer:
Scrat [10]3 years ago
4 0

Answer: the answer is attached

Step-by-step explanation:

2y/y-3. 4y-12/2y+6

Download docx
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8(C - 9) = 6(20 - 12) - 40​
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Answer:

All Real Numbers

Step-by-step explanation:

Any value of <em>c</em> makes the equation true.

Interval Notation

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⦁ The family likes peaches and other fruits. The local farmer’s market is selling fresh peaches at a discounted price. If the fa
sasho [114]

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98

The answer is 98 because since the farmer's market sold 35% of its produce and has 280 peaches remaining you have to turn 35% into a decimal which is 0.35 then multiply that times 280 and get 98.

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How to determine proofs
wariber [46]

Answer:

check footprints and then check finger print the finger print pour powder on it and Trace

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3 years ago
The cost of the apple is expressed as y = 12x + 6, where x is the number of apples. Find the cost of dozen apples.
astra-53 [7]

Answer:

<h2>             150</h2>

Step-by-step explanation:

x = dozen = 12

12x + 6 = 12•12 + 6 = 144 + 6 = 150

8 0
3 years ago
Find the volume of the solid obtained by rotating the region bounded by y=4x and y=2sqrt(x) about the line x=6.
Pie

Check the picture below.

so by graphing those two, we get that little section in gray as you see there, now, x = 6  is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.

y=4x\implies \cfrac{y}{4}=x\qquad \qquad y=2\sqrt{x}\implies \cfrac{y^2}{4}=x~\hfill \begin{cases} \cfrac{y}{4}=x\\\\ \cfrac{y^2}{4}=x \end{cases}

the way I get the radii is by using the "area under the curve" way, namely, I use it to get R² once and again to get r² and using each time the axis of rotation as one of my functions, in this case the axis of rotation will be f(x), and to get R² will use the "farthest from the axis of rotation" radius, and for r² the "closest to the axis of rotation".

\stackrel{R}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y^2}{4}}}\qquad \qquad \stackrel{r}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y}{4}}}~\hfill \stackrel{R^2}{\left( 6-\cfrac{y^2}{4} \right)^2}-\stackrel{r^2}{\left( 6-\cfrac{y}{4} \right)^2} \\\\\\ \stackrel{\textit{doing a binomial expansion and simplification}}{3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16}}

now, both lines if do an equation on where they meet or where one equals the other, we'd get the values for y = 0 and y = 1, not surprisingly in the picture.

\displaystyle\pi \int_0^1\left( 3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16} \right)dy\implies \pi \left( \left. \cfrac{3y^2}{2} \right]_0^1-\left. y^3\cfrac{}{} \right]_0^1-\left. \cfrac{y^3}{48}\right]_0^1+\left. \cfrac{y^5}{80} \right]_0^1 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{59\pi }{120}~\hfill

7 0
2 years ago
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