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dangina [55]
2 years ago
14

Jaxson bought snacks for his team's practice. He bought a bag of oranges for $2.91

Mathematics
1 answer:
Zinaida [17]2 years ago
3 0

Answer:

1.19

Step-by-step explanation:

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babunello [35]

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I think that the answer is A.

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(2x+3y) + (5x+3y) = 4 +-8
Artist 52 [7]

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6 0
3 years ago
6÷2(1+2)= _____ whats the answer to this
GalinKa [24]

Answer:

9

Step-by-step explanation:

You do parenthesis first so 1+2=3

Then you do division 6 divided by 2=3

Then you will multiply 3 by 3 so 9

This is known as pemdas

P=parenthesis

E=exponents

MD=multiplication and division you do it left to right so if division is in front of multiplication then you do division

As=Addition and subtraction is also left to right

8 0
2 years ago
Read 2 more answers
A group of eight golfers paid $430 to play a round of golf . Of the golfers one was a member and 7 were not what is the cost for
charle [14.2K]

Answer:

Incomplete question

Complete question;

A group of eight golfers paid $430 to play a round of golf . Of the golfers one was a member and 7 were not.

Another group of golfers consists of two members and one nonmember. They paid a total of $75. What is the cost for a member to play a round of golf, and what is the cost for a nonmember?

Answer: X = $82.695 for members

Y = $49.615 for non members

Step-by-step explanation:

Let's use X to denote members and Y for non-members.

Therefore, amount paid by one member to play + amount paid by 7 non-members to play = 430

X + 7Y = 430. . .1

Amount paid by 2 members to play + amount paid by one non-member to play = 215

2X + Y = 215. . .2

Solving both equations simultaneously

X+7Y = 430

2X +Y = 215

Therefore, from eqn 1. X = 430-7y

Substituting this into wan 2 gives

2(430-7Y) + Y = 215

860-14Y + Y = 215

860-215 =13Y

645 = 13y

Y = 49.615

Therefore substituting Y = 49.615 into any equation above

X + 7(49.615) = 430

X = 430-347.05

X = 82.695

3 0
3 years ago
HELP HELP!!
Goryan [66]

Answer:

0

Step-by-step explanation:

p_1:~~y = x^2+2\\p_2:~~y = 3x^2+2\\ \\ V{p_1} = \Big(-\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) = \Big(-\dfrac{0}{2}, -\dfrac{0^2-4\cdot 2}{4}\Big) = \Big(0,2\Big) \\ \\ Vp_2 = \Big(x_V, -\dfrac{\Delta}{4a}\Big) = \Big(0, -\dfrac{0^2-4\cdot 3 \cdot 2}{4\cdot 3}\Big) = \Big(0,2\Big) \\ \\ \\ \text{The distance is }0,~~\text{Because the vertices are equal.}

7 0
2 years ago
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