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4 years ago

A pizza stand offers both hand-tossed and pan pizza. Customers can add any combination of the six available toppings.

Mathematics

1 answer:

damaskus [11]4 years ago

4 0

Answer:

a)14 or b)74

Step-by-step explanation:

I'm assuming the question is how many combinations are there.

A) If toppings can't stack it would be 14, since there are 6 toppings for 2 pies and also you can go without toppings. or *2

B) If toppings can stack, then there can be 74 from being able to stack 6 toppings on one pizza and then two pizzas without it or (6^2)*2+2

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2(10r-7)-2(1+10r)=-r+2r

DerKrebs [107]

Answer:

r = -16

Step-by-step explanation:

We will use "distributive" property and "adding like terms" concepts in solving for the unknown variable "r".

The distributive property is:

a(b+c) = ab + ac

2(10r-7) = 20r - 14

also

-2(1+10r) = -2 - 20r

So, our equation becomes:

$2(10r-7)-2(1+10r)=-r+2r\\20r-14-2-20r=-r+2r$

Now we add like terms and move variables (r) to one side and numbers to another. Like terms mean:

3r + 5r = (3+5)r = 8r

So, we have:

$20r-14-2-20r=-r+2r\\-16=r\\r=-16$

So,

the value of r is -16

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3 years ago

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rjkz [21]

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3 years ago

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nirvana33 [79]

Answer:

Step-by-step explanation:

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3 years ago

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vitfil [10]

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Find (f∘g)(x).<br> f(x)=5x+1<br> g(x)=2x<br> Write your answer as a polynomial in simplest form.

vladimir2022 [97]

(f o g)=f(g(x))
just sub
f(g(x))
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4 years ago