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Kitty [74]
3 years ago
8

A lunch shop offers 2 kinds of soups, 3 kinds of sandwiches, and 3 kinds of beverages. How many combinations of one soup, one sa

ndwich, and one beverage are possible?
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

there are 18 combinations possible

Step-by-step explanation:

Lets say the soups are Mushroom(M) and Chicken (C)  

The sandwhiches are Tuna (T), Salad (S) and Ham (H)  

The drinks are Water (W), Pineapple Juice (P) and raspberry shake (R)  

The only arrangements that you can get are  

MTW  

MTP  

MTR  

MSW  

MSP  

MSR  

MHW  

MHP  

MHR  

CTW  

CTP  

CTR  

CSW  

CSP  

CSR  

CHW  

CHP  

CHR  

Total 18

i got this from a guy named J.J on yahoo

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Solve the equation using the quadratic formula
amm1812

Answer:

Step-by-step explanation:

Rewrite this quadratic equation in standard form:  2n^2 + 3n + 54 = 0.  Identify the coefficients of the n terms:  they are 2, 3, 54.

Find the discriminant b^2 - 4ac:  It is 3^2 - 4(2)(54), or -423.  The negative sign tells us that this quadratic has two unequal, complex roots, which are:

     -(3) ± i√423        -3 ± i√423

n = ------------------- = ------------------

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3 years ago
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Keith_Richards [23]

Answer:

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hope it helps

Step-by-step explanation:

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4 0
3 years ago
Find the values of y = p(x) = √x for<br> x = 0, 1.44, 2.25, 3.24, 4.41, 5.29
zhenek [66]
P(x) = √x


for x = 0 → √0 = 0 and p(0) = 0
for x = 1.44 → √1.44 =1.2  and p(1.44) = 1.2
for x = 2.25 → √2.25 = 1.5 and p(2.25) = 1.5
for x = 3.24 → √3.24 = 1.8 and p(3.24) = 1.8
for x = 4.41 → √4.41=  2.1 and p(4.41) = 2.1
for x = 5.29 → √5.29 = 2.3 and p(5.29) = 2.3

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if a right circular cone has a volume of 15 cubic ft and a radius of 3 ft, what is the height of the cone?
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Volume of cone=1/3 times height times pir^2
v=15
r=3
15=1/3 hpi3^2
15=1/3hpi9
15=9/3hpi
15=3hpi
divide 3
5=hpi
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1.59 ft
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just olya [345]
4 is 80% of 5, because 1 is worth 20% 
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3 years ago
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