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g100num [7]
3 years ago
12

You have a bag with 30 fireballs and another bag with 42 jolly ranchers. For a party, you want to repackage the candy into small

er bags, and you want each bag to have the same number of fireballs and jolly ranchers and. How many bags will you be able to make without any leftover?
Mathematics
2 answers:
Andrew [12]3 years ago
7 0
For every 5 fireball, there is 7 jolly ranchers
Lady_Fox [76]3 years ago
4 0
ANSWER: 13




I just added and the dividend lol
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I graphed this line below.

Start with an open dot on 7.

The reason we use an open dot is because x

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Finally, state your answer in set notation if possible.

It is read as {x: x > 7}.

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frosja888 [35]

Answer:

130

Step-by-step explanation:

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3 years ago
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John took a quiz there are 25 questions, I got a 76% how many did John get right and how many did he get wrong
FinnZ [79.3K]

Answer:

6 wrong.

Step-by-step explanation:

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Hope this helps plz hit the crown :D

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3 years ago
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Evaluate 5 + 6 · 2 – 8 ÷ 4 + 7 using the correct order of operations
Natali5045456 [20]
Do solve this you use PEMDAS. This stands for:
Parenthesis
Exponents
Multiplication
Division
Addition
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Because there are no parentheses or exponents in this equation, we go straight to the division.

1) 8 divided by 4 equals two so we rewrite the equation. 5+6 x 2 - 2 + 7.

2) Next in order of pemdas is multiplication. Yes I know that in pemdas multiplication does come before division but it doesn’t matter in this problem which you do first. So now we do 6 x2 =12. We then rewrite the equation 5+12-2+7.

3) lastly, addition and subtraction are grouped together and you can just do it in order of the problem. So we do 5+12=17. 17-2=15. And 15+7=22.

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8 0
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Can someone please help me factor 6x^2 - 9x+42
frutty [35]

Answer:

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Step-by-step explanation:

6x^2-9x+42

All three terms have a common factor of 3

3(2x^2-3x+14)

Now let's focus on 2x^2-3x+14 and bring down the factor 3 later

so a=2

b=-3

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Let's try to find two factors for ac that multiply to be a*c and add up to be b.

ac=28

b=-3

-----

ac=7(4)=14(2)=8(2)

Even if I made these pairs with both negatives nothing would give me -3

So you can only go as far as 3(2x^2-3x+14)

Here is another thing to help you if you have ax^2+bx+c and b^2-4ac<0 then it can't be factored (over reals)

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