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Olenka [21]
3 years ago
5

Jake needs to put gas in the delivery van regular gasoline is 3.20 per gallon he only had a $20 bill on him how many gallons can

he buy? Plz help someone
Mathematics
2 answers:
kumpel [21]3 years ago
5 0
If we want to see how much gas Jake can buy, we have to divide. We know that gas is $3.20 per gallon and he has $20. We have to divide 20 and 3.20, so we get 20/3.20 = 6.25. Jake can get 6.25 gallons of gas. Or 6 1/4 gallons of gas if you want the answer with a fraction.
Fed [463]3 years ago
4 0
Your max amount for this is $20. Let's say x represents the number of gallons.

20 ≥ 3.2x
x = 6.25 gallons
If they're asking for integers or whole numbers, you'd round down to 6 because you only have enough money for 6 full gallons. If you rounded up to 7, you wouldn't be able to fill it up completely because then you'd be over your $20.00 budget for gas.
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What's the slope between (-2, 8) and (4, 20)
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
Please help me with this question
azamat
It is 90 your welcome
7 0
3 years ago
You have a wire that is 38 cm long. you wish to cut it into two pieces. one piece will be bent into the shape of a square. the o
erica [24]
Circumference + perimeter =38
x = circumference of circle
find r in terms of x
2\pir=x
r=\frac{x}{2 \pi }
Area=\pir^2
so..
A=\pi(\frac{x}{2 \pi })^2
=\pi(\frac{x^{2} }{4 \pi ^{2} })
=\frac{x^{2} }{4 \pi }

(38-x) is perimeter
then 
\frac{38-x}{4}
(\frac{38-x}{4})^2
\frac{x^{2} }{4 \pi } + (\frac{38-x}{4})^2
then you graph it and it equals
16.716 cm the circumference of the circle
6 0
3 years ago
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
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