X÷4+9=12 plzzzzzzzzzzzzzzz help step by step

1 answer:

6 0

So we are trying to solve for x so in order to do that we need to get x by itself on one side of the equal sign

first we are going to combine like terms on each side of the equal side

so on the left side we see 4+9 these are booth whole numbers not attached to a variable so we can simplify this by adding these two values together making our equation: x/13 = 12

our x is still not by itself so in order to get it by itself we need to get rid of that 13

we need that x is being divided by 13 so in order to get rid a divided value we have to do the opposite to both sides - multiply by 13 on each side!

x/13 x 13 = x
12 x 13 = 156

so our equation is now:
x = 156

is our x by itself on one side of the equal sign? yes! so our x value is 156

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victus00 [196]

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If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

Ilya [14]

This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
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34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34 + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

$\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}$

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}$

and now let's use those 2 found folks, to get the 21st term,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104$

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Write the decimal that 1 tenth of 0.9

Viefleur [7K]

0.09

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