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Lynna [10]
4 years ago
15

If 63

%5C%3A%20then%20%5C%3A%20k%20%3D%20" id="TexFormula1" title=" {63}^{2} = 1 + 3 + 5 + ... + k \: then \: k = " alt=" {63}^{2} = 1 + 3 + 5 + ... + k \: then \: k = " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Scilla [17]4 years ago
3 0

Answer:

125

Step-by-step explanation:

let's see if we can find a pattern

1 = 1 = 1^2

1 + 3 = 4 = 2^2

1 + 3 + 5 = 9 = 3^2

we are starting to see a pattern here, when we add another term, we find the next integer squared.

we can find this last number in our sequence with 2x-1.

so in the next line we should find 4^2. if we plug 4 into 2x-1 we would expect the last term in the sum to be 7.

1 + 3 + 5 + 7 = 16 = 4^2

so 2*63 -1 = 126 - 1 = 125

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3 years ago
The glee club is having their annual Spring performance. The equation 5a+2s=48 represents the money collected on the first night
V125BC [204]
First, put them directly above each other.
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3a+2s=32

You need to combine these, but in order to do so, we want to get rid of one of the letters. The S's line up very well, and all we need to do is multiply the bottom equation by a negative 1. This will look like this,
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Divide both sides by 2 to get a by itself.

a=8, is your answer!

If it asked for s, you would put 8 into whichever equation seems easier for you, and then its just simple crunching it out.

Hope this helped!
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3 years ago
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Answer:

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Step-by-step explanation:

Change to improper fractions format

1 6/10 becomes 16/10

2 7/10 becomes 27/10

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3 0
3 years ago
Read 2 more answers
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