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ElenaW [278]
3 years ago
14

ellen has two quarters one dime one nickel how many different amounts of money can she make using one or more of these coins

Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0
Ellen can make 10 different amounts.
5102525 + 5 = 3025 + 10 = 3525 + 10 + 5 = 4025 + 25 = 5025 + 25 + 5 = 5525 + 25 + 10 = 6025 + 25 + 10 + 5 = 65
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The function d(x)=1375-110x represents the distance (in miles) a high speed train is from its destination after x hours.
Verizon [17]

Answer:

12.5 hours

Step-by-step explanation:

The function d(x) = 1375 -110x represents the distance in miles a high-speed train is from its destination after x hours of travel.

So, when the train reaches its destination then d(x) will become zero miles.

Therefore, 0 = 1375 - 110x

⇒ x = 12.5 hours.

So, the train will travel 12.5 hours to reach its destination. ( Answer )

3 0
3 years ago
The Holcombe family went out to dinner
Jet001 [13]

Answer:

$13.77 per person in the family

Step-by-step explanation:

Cost of the meal = $54.65

Sales tax rate = 8%

Sales tax amount = 8% of $54.65

= 8/100 × 54.65

= 0.08 × 54.65

= $4.372

Tip = 18% of $54.65

= 18/100 × 54.65

= 0.18 × 54.65

= $9.837

Total cost of the dinner = Cost of the meal + Sales tax amount + Tip

= $54.65 + $4.372 + $9.837

= $68.859

There are 5 people in the Holcombe family

What was the cost per person for the meal, including sales tax and tip?

cost per person for the meal, including sales tax and tip = Total cost of the dinner / total number of people in the family

= $68.859 / 5

= $13.7718

Approximately

$13.77 per person in the family

5 0
3 years ago
Willl mark brainlyist
devlian [24]

Answer: I would think C, 1/11, but think about it though. I don’t know if it is right.

6 0
3 years ago
The vertex of this angle is: <br> a) T<br> b) R<br> c) S
Dvinal [7]

Answer:

The vertex of angle is S.

Hope this helps!!

7 0
4 years ago
Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.
jek_recluse [69]

Answer:

\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that \exists x \in X so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that \exists x \in X so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that \forall x \in X none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)

\mathbf{e)}

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

This means that for each of the statements C, F and D there is an element from the domain X so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

5 0
4 years ago
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