Ask question

Ask question

All categories

3 years ago

7

0.6 rounded to the nearest tenth

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

1.0

Step-by-step explanation:

The 6 is bigger than 5, so that bups the number up to 1.0.

You might be interested in

Help pls!!!!!!!!!!!!!!!!!

const2013 [10]

Answer: just turn them into decimals and use a calculator then turn them back into fractions in simplified form

Step-by-step explanation:

7 0

3 years ago

What is x² + x + 49 on 2020 edg?

Verdich [7]

Answer:

259

Step-by-step explanation:

6 0

3 years ago

A color printer prints 31 pages in 9 minutes. How many pages does it print per minute?

djyliett [7]

According to my calculations, it would take approximately 3.4 (being in decimal form) pages PER minute or Just 3 minuets being rounded to the nearest tenth. If I am I correct please correct me. I used a calculator.

7 0

2 years ago

If BC=CD=94 and AD=59, what is AB?

Novay_Z [31]

Answer: I do not know

Step-by-step explanation:

7 0

3 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

3 years ago