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3 years ago

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

1 answer:

7 0

Use math induction proof method will be usefully

for n=1 will get

1^2 = 1(6*1^2 -3*1 -1)/2
1 = 1(6-4)/2
1 = 1(2)/2
1 = 1

suppose true for n=k and prove for k = k+1

hope this will help you

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Answer:

1. Translation: g(x) = f(x-a)+b.

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3. Reflection around x-axis: k(x) = -f(x)

4. Rotation of 90° : $R_{90} (x,y)=(-y,x)$

5. Rotation of 180° : $R_{180} (x,y)=(-x,-y)$ .

6. Rotation of 270° : $R_{180} (x,y)=(y,-x)$ .

Step-by-step explanation:

Let us assume that the transformations namely translation, reflection are applied to a function f(x) and the rotation is applied to the point ( x,y ).

So, according to the options:

We know that 'translation moves the image in horizontal and vertical direction'.

1. As we have to translate the function f(x) 'a' units to the right and 'b' units up. So, the new form of the function becomes g(x) = f(x-a)+b.

Further, we know that 'reflection means to flip the image around a line'.

2. As, we have to reflect the function f(x) around y-axis. The new form of the function is h(x) = f(-x).

3. As, we have to reflect the function f(x) around x-axis. The new form of the function is k(x) = -f(x).

Since, 'rotation turns the image around a point to a certain degree'.

4. As, we have to rotate ( x,y ) counter-clockwise to 90° about the origin, the new form of the function is $R_{90} (x,y)=(-y,x)$ .

5. As, we have to rotate ( x,y ) counter-clockwise to 180° about the origin, the new form of the function is $R_{180} (x,y)=(-x,-y)$ .

6. As, we have to rotate ( x,y ) counter-clockwise to 270° about the origin, the new form of the function is $R_{180} (x,y)=(y,-x)$ .

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2 years ago

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