Answer:
ΔA'B'C' is a reduction of ΔABC and ΔA'B'C' is similar to ΔABC.
Step-by-step explanation:
It is given that the triangle ABC is dilated to produce triangle A'B'C' with scale factor 3/4.
If a figure is dilated then preimage and image are similar.
If scale factor is between 0 to 1, then preimage is reduction of image.
If scale factor is more that 1, then preimage is enlargement of image.
If scale factor is 1, then preimage is congruent to the image.
We know that
So,
Therefore, the ΔA'B'C' is a reduction of ΔABC and ΔA'B'C' is similar to ΔABC.
Base case: if <em>n</em> = 1, then
1² - 1 = 0
which is even.
Induction hypothesis: assume the statement is true for <em>n</em> = <em>k</em>, namely that <em>k</em> ² - <em>k</em> is even. This means that <em>k</em> ² - <em>k</em> = 2<em>m</em> for some integer <em>m</em>.
Induction step: show that the assumption implies (<em>k</em> + 1)² - (<em>k</em> + 1) is also even. We have
(<em>k</em> + 1)² - (<em>k</em> + 1) = <em>k</em> ² + 2<em>k</em> + 1 - <em>k</em> - 1
… = (<em>k</em> ² - <em>k</em>) + 2<em>k</em>
… = 2<em>m</em> + 2<em>k</em>
… = 2 (<em>m</em> + <em>k</em>)
which is clearly even. QED
Answer:
Step-by-step explanation:
Hence, the surface area of a triangular pyramid is 140 square units.
It depends how you want to solve it
you can plug it in on a calculator and It will show you how the equation looks on a graph( need graphing calculator).
Answer:
(Sin A + Cos A)/Sin A. Cos A
Step-by-step explanation:
As we know
Sec A = 1/Cos A
and Cosec A = 1/Sin A
Given Equation
Sec A + Cosec A
Substituting the given values, we get -
1/cos A + 1/Sin A
(Sin A + Cos A)/Sin A. Cos A
