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zzz [600]
3 years ago
13

Helppppp distance time graphhhhhh

Mathematics
1 answer:
stepladder [879]3 years ago
3 0

Answer:

Ralph walked to Sherwood at a constant speed.

Step-by-step explanation:

Any horizontal lines on a distance time graph represent a moment where you are still, so that rules out graph 4 and graph 3. Because Ralph is at a constant speed, the gradient of the line won’t change, meaning the answer has to be Ralph walked to Sherwood at a constant speed.

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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

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Answer:

an equation of the first degree in any number of variables.

Step-by-step explanation:

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4 years ago
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MatroZZZ [7]

Answer:

Step-by-step explanation:

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3 years ago
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WILL ANSWER FIRST RIGHT ANSWER AS BRAINIEST
Ket [755]

Answer:

8

Step-by-step explanation:

Considering the given value of sum of interior angles with exclusion of one to be 1070 then we can write it as

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lakkis [162]

Answer:

the answer is 2.89

Step-by-step explanation:

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