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igomit [66]
3 years ago
14

A value meal package at Ron's Subs consists of a drink, a sandwich, and a bag of chips. There are 4 types of drinks to choose fr

om, 3 types of sandwiches, and 3 types of chips. How many different value meal packages are possible?
Mathematics
2 answers:
Temka [501]3 years ago
8 0

To answer this question, multiply all given numbers together.

4*3*3

12*3

36

36 different value meal packages are possible.

Hope this helps!

dedylja [7]3 years ago
6 0

Answer:

There are 36 different combinations.

Step-by-step explanation:

A value meal (V) is formed by 1 drink, 1 sandwich and 1 bag of chips. There are:

  • 4 types of drinks (d)
  • 3 types of sandwiches (s)
  • 3 types of chips (c)

To find all the possible combinations we use the following expression.

V = d × s × c

V = 4 × 3 × 3

V = 36

There are 36 different combinations.

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Which angle number is a vertical angle to
VMariaS [17]

Answer:

Angle 7 or angle DHG

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5 0
3 years ago
WILL MARK BRAINLIEST <br><br> what is the domain of (f/g) (x)?
tatyana61 [14]

Given that f(x) = \sqrt{7-x} and g(x) = \sqrt{x + 2}, we can say the following:

\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}


Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).


Thus, let's set what is under both square roots to be greater than 0:

\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7

\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2


Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}


Now, let's look back at the function entirely, which is:

\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}

Since \sqrt{x + 2} is on the bottom of the fraction, we must say that \sqrt{x + 2} \neq 0, since the denominator can't equal 0. Thus, we must exclude \sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2 from the domain.


Thus, our answer is Choice C, or \boxed{ \{ x | -2 < x \leq 7 \}}.


<em>If you are wondering why the choices begin with the x | symbol, it is because this is a way of representing that x lies within a particular set.</em>

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