Ask question

Ask question

All categories

4 years ago

14

HELP! PLEASE! which graph represents a function? How do you know?

2 answers:

Andreas93 [3]4 years ago

5 0

I agree the answer is D

WITCHER [35]4 years ago

4 0

Any graph will represent a function if it passes the vertical line test. A graph is a function if for any vertical line drawn it touches the graph at only one point.In other words for every x there is only one y value.

In the given graphs if a vertical line is drawn in the graphs A,B it will touch the graph at two points and hence they are not unctions.Option c s a vertical line that is for one x value there are more than one value of y and so is not a function.Option D will represent a function as any vertical line drawn will touch the graph at only one point.

OPtion D.

You might be interested in

A,B,C,D? I would thank you a lot...

alekssr [168]

Answer:

here, ABCDEFGHIJKLMNOPQRSTUVWXYZ

Step-by-step explanation:

7 0

3 years ago

X + 318 = 718 x = x – 212 = 634 x =<br> x – 3.23 = 0.77 x =

kogti [31]

Did you try photomath

6 0

3 years ago

Sam loves basketball! he can make the shot 65% of the time. If he takes 30 shots how many times will he make it?

UkoKoshka [18]

Answer:

20

Step-by-step explanation:

30 * 0.65 = 19.5

Round off the number to 20

7 0

3 years ago

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.

Arisa [49]

Answer: $S_n=5(1-\dfrac{1}{n+1})$ ; 5

Step-by-step explanation:

Given series : $[\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]$

Sum of series = $S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]$

Consider $\dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}$

$=\dfrac{1}{n}-\dfrac{1}{n+1}$

⇒ $S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]$

Put values of n= 1,2,3,4,5,.....n

⇒ $S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})$

All terms get cancel but First and last terms left behind.

⇒ $S_n=5(1-\dfrac{1}{n+1})$

Formula for the nth partial sum of the series :

$S_n=5(1-\dfrac{1}{n+1})$

Also, $\lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})$

$=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5$

4 0

3 years ago

Which number sequence follows the rule add 4 starting from 2?

Kaylis [27]

2 6 10 14 18 22 26 30 and so on

6 0

2 years ago