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3 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]3 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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Levart [38]

X=7.
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Y=-9
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2 years ago

14. Let R^2 have inner product defined by ((x1,x2), (y,, y2)) 4x1y1 +9x2y2 A. Determine the norm of (-1,2) in this space B. Dete

Vitek1552 [10]

The norm of a vector $\vec x$ is equal to the square root of the inner product of $\vec x$ with itself.

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3 years ago

Maria, Bill, and Eric have a total of $95 in their wallets. Maria has 10 less than Bill. Eric has 3 times what Bill has. How muc

BaLLatris [955]

Answer:

$\boxed{Maria \: got \: \$11} \\ \boxed{Bill \: has \: \$21} \\ \boxed{ Eric \: has \: \$63}$

Step-by-step explanation:

$let \: maria \: be \to \: m \\ let \: eric \: be \to \: e \\let \: bill \: be \to \: b \\then : m =b - 10 \: and \: e = 3b \\ but : m + b + e = 95 \\ hence \to \: (b - 10) + b + (3b) = 95 \\ 2b - 10 + 3b = 95 \\ 5b = 95 + 10 \\ 5b = 105 \\ \boxed{b = 21} \\ m = b - 10 = 21 - 10 \\ \boxed{ m = 11} \\ e = 3b = 3(21) \\ \boxed{e= 63}$

♨Rage♨

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Svet_ta [14]

Answer:

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Step-by-step explanation:

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Mars2501 [29]

Answer:

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Step-by-step explanation:

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I hope this makes sense lol, I pretty much guessed the answered so don't expect a proper explanation. :'(

Anyways, there is a screenshot attached that might help.

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2 years ago