Ask question

Ask question

All categories

3 years ago

14

Answer this please!!! Need help now!

1 answer:

Simora [160]3 years ago

6 0

It would be D...
not quite sure, cause most of these are not a function

You might be interested in

On a school trip to a theme park, four busses each carry 105 students. 35% of the students are bringing their own lunch. 3/7 of

vodka [1.7K]

Answer:

We have a lunch pack for her lunch

3 0

3 years ago

Ellie is making a batch of pancakes

lys-0071 [83]

Answer:

Ellie needs 1.2 or 1 1/5 cups of flour (as 2/10 simplifies to 1/5).

Please mark me brainliest if this helped!

7 0

3 years ago

Read 2 more answers

Simplify 3*a*2*b<br> *=multiply

morpeh [17]

Answer:

3 × a × 2 × b = 3a × 2b

7 0

4 years ago

Bob’s gas tank is 1/2 full. After he buys 3 gallons of gas, it is 2/3 full. How many gallons can Bob’s tank hold?

ankoles [38]

The answer is his tank can hold 15 gallons i think

8 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