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JulsSmile [24]
3 years ago
6

20cm cake serves 12 people costs 13.50 per cake and 25cm cake serves 20 people costs 18.75 . There will be 57 guests and want to

make sure everyone gets a slice . Want to pay as little for the cakes and want the cakes to be the same size how much will I have to pay for the cakes
Mathematics
1 answer:
nadya68 [22]3 years ago
7 0
So,

If you do some simple division, you will find how much each cm costs in each cake.
20cm cake = 67.5 cents/cm
25cm cake = 93.75 cents/cm

Assuming that you cannot order fractions of cakes, the best deal would be to order 5 of the 20cm cakes.
Cost = 13.50 * 5 = $67.50

You will have to pay $67.50 for the cakes if you want the best deal.
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luda_lava [24]
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7 0
2 years ago
(Geometry)
tatiyna

Answer:

A)Isosceles Triangle

B)Scalene Triangle

C)Equilateral Triangle

D)Isosceles Triangle

E)Scalene Triangle

F)Isosceles Triangle

Step-by-step explanation:

Helped me out with the triangles on the bottom but remember

all sides diffrent= Scalene Triangle

All sides the same = Equilateral Triangle

Only 2 sides the same = Isosceles Triangle

5 0
3 years ago
F(x)<br> = x + 2<br> g(x) = 3x + 3<br> Find (f•g)(-6)
Lunna [17]

Answer:

(-4*-15) = 60

4 0
3 years ago
Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
3 years ago
What is 3 cubed by 3 over 9 plus 12
saw5 [17]
Use your pemdas! so you do the parentheses first
then move on to the exponent.
then multiply.
the answer would be whatever 9 times (37/3) is

7 0
3 years ago
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