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3 years ago

A sprinter ran a race in 11 seconds. What is the domain of this graph?

Mathematics

1 answer:

gtnhenbr [62]3 years ago

4 0

Domain is defined as the all possible values of the variable x for which the function is defined.

Now, for the given graph the x-axis represents the time in seconds.

So, we need to find all the possible values of time for which the distance is covered or for which the function is defined.

Given that : The sprinter ran a race in 11 seconds.

And the initial time is 0 seconds

So, The value of t lies between 0 to 11

Thus, Domain : All real numbers 0 to 11

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2 years ago

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

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Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

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3 years ago