"The grid shows Figure Q and its image Figure Q’ after a transformation. Which transformation was applied on Figure Q?

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

8 0

The answer is C because if you turned Figure Q 90 degrees clockwise, it matches Q', thus a counterclockwise rotation of 270 degrees around the origin had to have been applied.

A is disqualified because if it had been rotated 90 degrees counterclockwise, it would be in the top left quadrant.

B is disqualified because if it was rotated 180 degrees about the origin, it would be in the bottom left quadrant.

D is disqualified because if rotated 270 degrees about the origin it would be in the top left quadrant, same as answer A.

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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
34 + d + d + d + d + d + d
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,

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

3 0

2 years ago

Which expression is equivalent to (2mn)4/6m^-3n^-2

Scrat [10]

Answer is the option 1
Because the denominator is 6/