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Lena [83]
3 years ago
6

A grocery store normally sells lemonade for $3.50 per bottle grocery store is currently having a sale on lemonade which advise s

ix bottles for only 1350 how much cheaper it is eliminated on sale per bottle compared to the normal price per bottle
Mathematics
1 answer:
anygoal [31]3 years ago
5 0

Answer:

7.50 cheaper than buying individually

Step-by-step explanation:

13.50/6=2.25

3.50*6=21.00

21.00-13.50= 7.50

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PLEASE HELP!!!
kramer

The arc is defined by an angle of 1.05 radians, and a length of 1.05 units.

<h3 /><h3>How to get the angle in radians?</h3>

Remember the relation:

180° = 3.14 radians.

Then:

60° = (60°/180°)*3.14 rad = 1.05 rad.

Now, for a circle of radius R, an arc defined by an angle θ has a length:

L = θ*R

Then the length of this arc is:

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2 years ago
I need help with this question
Wittaler [7]

Answer:

A=W, B=X, C=Y, D=Z, AB=WX, BC=XY, CD=YZ, AD=WZ

(The second answer down)

Step-by-step explanation:

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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

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Answer:

area : pi*r^2

pi * 20^2

400pi m^2

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