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How much lemonade will each person get if 4 people share 1/2 gallon equally?

Charra [1.4K]

I’m pretty sure the answer would be 8 if u cut 4 in half 4 times

3 0

3 years ago

A cylinder has a height of 4 meters and a radius of 1.5 meters. What is the approximate radius of a sphere that has the same sur

Mama L [17]

If the surface area of the cylinder is 12π square meters. Then the radius of the sphere will be 1.7 meters.

<h3>What is a cylinder?</h3>

A cylinder is a closed solid that has two parallel circular bases connected by a curved surface.

A cylinder has a height of 4 meters and a radius of 1.5 meters.

Then the approximate radius of a sphere that has the same surface area as the cylinder.

We know that the Surface area of the cylinder will be is given as

Surface area = 2πrh

Surface area = 2 x π x 1.5 x 4

Surface area = 12π square meters

Then we have

Surface area of sphere = surface area of cylinder

4πr² = 12π

r² = 3

r = 1.7 meters

More about the cylinder link is given below.

brainly.com/question/3692256

#SPJ1

4 0

2 years ago

A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain

AysviL [449]

Answer:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

Step-by-step explanation:

For this case we have the following value:

$R = 0.0498 \frac{Kpa}{Km}$

We can convert this first to $\frac{Kpa}{m}$ like this:

$R=0.0498 \frac{Kpa}{Km} *\frac{1km}{1000m}=0.0000498 \frac{Kpa}{m}$

Now we use the fact the the pressure is defined as $P =\frac{F}{A}$ , whre P is the pressure, F the force and A the area, so then $Kpa= \frac{KN}{m^2}$ and then we can replace this:

$R=0.0000498 \frac{KN}{m^3}$

Now from definition of work we know that $W= Fd$ where W is the work, F the force and d the distance, so then is equivalent $KN =\frac{KJ}{m}$

And if we replace this into the equation we got:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

4 0

3 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

(write the from for each number and give the value of the underline digit) 2.(3)00

Mkey [24]

The answer is three hundred

7 0

3 years ago