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VashaNatasha [74]
3 years ago
6

Help me plzzzzzz plz

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0
Undefined. Vertical lines are always undefined. Hope that helps!
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If you horizontally stretch the quadratic parent function, f(x) = x2, by a factor
Brilliant_brown [7]

Answer:

Step-by-step explanation:

If you horizontally stretch the quadratic parent function, f(x) = x2, by a factor

of 3, what is the equation of the new function?

A. g(x) = 1/3x2

B. g(x) = 3x2

C. g(x) = (3x)2

D. g(x) = (1/3x)2

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3 years ago
31 x 01111111111111111111111111
dybincka [34]
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A shape is made up of 8 right triangles of equal size. each right triangle has a base of 4cinches and a height of 15 inches. wha
Basile [38]

Answer:

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Step-by-step explanation:

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3 years ago
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If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
Please help me on 7&amp;8 and please show your work :)
Llana [10]
7. 6/12 or 1/2 | 8. 5/12

5 0
3 years ago
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