All categories

3 years ago

Whether the following relation represents a function. Use pencil and paper.

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

6 0

Answer: it’s yes because A function is a relationship between quantities

saveliy_v [14]3 years ago

5 0

The answer is O yes the relation does represent a function

You might be interested in

What number should be placed in the box to help complete the division calculation?

Mkey [24]

Answer:

I thought it was 9 but wasn't sure so looked it up and got the image below. Hope this helps.

P.s.

Please give me brainliest for my next rank.

3 0

3 years ago

Read 2 more answers

In circle O, central angle BOC has a measure of 115. Find the measure of arc BAC. In your final answer, Include all of your work

kakasveta [241]

Circumference you multiply and divide

7 0

3 years ago

Read 2 more answers

Need help Pronto! Will give braintliest.

PtichkaEL [24]

Answer:

c looks like the right answer.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

____ [38]

Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$ (with error 0.0001)

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

3 0

3 years ago

-36 4/9-(-10 2/9)-(18 2/9)

n200080 [17]

Answer:

-44.4444444444

Step-by-step explanation:

Calculator

5 0

4 years ago

Read 2 more answers