For the A parts just make all the second numbers negative example: if it were 5,1 make it (5,-1) and for the B part make all the first numbers a negative example: if it were 5,1 make it (-5,1)
Answer: 
Step-by-step explanation:
For this exercise it is important to know the definition of "Dilation".
A Dilation is defined as a transformation in which the Image (The figure obtained after the transformation) and the Pre-Image (The original figure before the transformation) have the same shape but have different sizes.
If the Dilation is centered at the origin, and knowing the scale factor of 
, you need to multiply each coordinate of the point T by the scale factor in order to find the coordinates of the Image T'.
Knowing that the point T has the following coordinates:
You get that the coordinates of the Image T' are the shown below:
Answer: I’m pretty sure it’s the 4th one.
Step-by-step explanation: because 1 and 2 have infinity solutions but 3 has no solution 4 comes up with x=-11
Can you give brainiest? :)
Answer:
0.125
Step-by-step explanation:
1/8=0.125
Step-by-step explanation:
LHS:
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}1+21+2+31+3+41+……………+8+91
Rationalizing the denominator, we get
\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)⇒(1+21×1−21−2)+(2+31×2−32−3)+(3+41×3−43−4)+⋯…+(8+91×8−98−9)
We know that,
\left(a^{2}-b^{2}\right)=(a+b)(a-b)(a2−b2)=(a+b)(a−b)
Now, on substituting the formula, we get,
=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\cdots \ldots \cdot \frac{(\sqrt{8}-\sqrt{9})}{8-9}=1−21−2+2−32−3+3−43−4+⋯…⋅8−9(8−9)
\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})⇒1+21+
