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3 years ago

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She sold each dozen for 4.50. how much did she earn from the eggs she sold?

1 answer:

anzhelika [568]3 years ago

5 0

CAN YOU BE MORE SPECIFIC? SO THAT THE QUESTION CAN GET ANSWERED

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It took Jaheim 2 hours to read the first 100 pages of a book what was his speed?

jeka94

If in 2 hours he read 200 pgs then in 1 hour will be 50 pgs

4 0

3 years ago

Read 2 more answers

A cereal company's cost, in thousands of dollars, is represented by the function C(x)=2x+4500 and its revenue, in thousands of d

jekas [21]

Given , revenue function : $R(x)= 5x$

Cost function $C(x) = 2x+4500$

Profit = Revenue - Cost

Hence, P = R(x) - C(x)

Total revenue for x = 12000 is $R(12000) = 5(12000) = 60000$

Total cost for x = 12000, $C(12000) = 2(12000)+4500 = 28500$

Profit = 60000-28500 = 31500

Answer = 31500

6 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:

<u><em>Question 1 a. i) Find the value of x.</em></u>

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

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:

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

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

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

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

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

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

<u>ii) Lower frustrum</u>

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

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

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

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

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

6 0

3 years ago

Simplify (5p+1.49)-(2p+6.49)<br> Thank you!

Xelga [282]

5p+1.49-2p-6.49
Combine like terms
3p+1.49-6.49
=3p-5

5 0

3 years ago

PLEASE HELP SOLVE!!!!!!!!<br><br> -72 < 8m < -32<br><br> -5r > -15 or r + 5 > 14

denis23 [38]

That's the first one

5 0

4 years ago