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3 years ago

What is 4/15 divided by 2/3?

Mathematics

1 answer:

BaLLatris [955]3 years ago

5 0

Answer: 0.4 or 2/5

Hope it helps!

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Question<br> (X-5y/y3)-1=

Ilia_Sergeevich [38]

Answer:

$x = y^3+5y$

Step-by-step explanation:

Complete question

$\frac{x - 5y}{y^3} - 1=0\\$

Required

Solve for x

We have:

$\frac{x - 5y}{y^3} - 1=0$

Collect like terms

$\frac{x - 5y}{y^3} = 1$

Multiply through by $y^3$

$x - 5y = y^3$

Make x the subject

$x = y^3+5y$

8 0

2 years ago

Which one of the following is an example of a repeating decimal A. 0.666666 B. 1.234545 C. 6.787878 D.0.123321

sweet [91]

The answer: A. 0.666666

4 0

3 years ago

Jesse has a 13.47-ounce bar of gold. He cuts 1.55 ounces off to make a piece of jewelry. How much gold is left in the bar?

telo118 [61]

Answer:

11.92 Ounces.

8 0

3 years ago

Which number sentence shows the correct placement of the decimal point in the product, based on the two factors?

snow_lady [41]

Answer:

Where are the number sentences?

6 0

3 years ago

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by y=√x and y=x^2. Find V either by sl

Ede4ka [16]

Answer:

The volume of the solid is $3\pi/10$

Step-by-step explanation:

In this case, the washer method seems to be easier and thus, it is the one I will use.

Since the rotation is around the y-axis we need to change de dependency of our variables to have $f(x)\rightarrow f(y)$ . Thus, our functions with $y$ as independent variable are:

$x=\sqrt{y}\\ x=y^2$

For the washer method, we need to find the area function, which is given by:

$A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]$

By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function $x=\sqrt{y}$ and the inner one by $x=y^2$ . Thus, the area function is:

$A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)$

Now we just need to integrate. The integration limits are easy to find by just solving the equation $\sqrt(y)=y^2$ , which has two solutions $y=0$ and $y=1$ . These are then, our integration limits.

$V=\pi\int_{0}^1 (y-y^4)dy\\ V=\pi (\int_{0}^1 ydy - \int_{0}^1 y^4dy)\\ V=\pi/2-\pi/5\\\boxed{V=3\pi/10}$

3 0

3 years ago