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ryzh [129]
4 years ago
10

How do I solve these problems that I took a picture of ?

Mathematics
2 answers:
fgiga [73]4 years ago
6 0
22. Step by step: 2x^2+5x-3=0
factor the left side
(2x-1)(x+3)=0
set the factors to equal zero
2x-1=0 and x+3=0
x=1/2           x=-3

23. Use same technique.
3x^2-13x-10=0
factor the left side once more
(3x+2)(x-5)=0
set factors to zero
3x+2=0 and x-5=0
x=-2/3          x=5
Inessa05 [86]4 years ago
4 0
Her work is correct! :)
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28 children will attend the birthday party if each child gets 2 party favors and there are 14 favors in each box how many boxes
dezoksy [38]
Hello!

So if each box has 14 favors, what we need to do is add until we have enough, or more than enough for each of the children that will be at the party to get two of them.

So to start, 14+14=28. There are 28 children going, but each child must have two favors, not one. So, we simply add on two more boxes. 28+14=42, 42+14=56. So, we need 56 favors for each child to get two. 

You could multiply to make this problem faster, but adding works out easier in my head. Sorry to drag it out!

So to make it simple, you need four boxes of favors for each child to get two.

Hope this helps, have an awesome day!
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