Question

Can someone please indicate the rise and run in the graph?
http://prntscr.com/3rwazw

Answer 1

see if that helps...

Answer 2

then how are we getting 3/1 as a slope for graph one

Answer 3

good question :) whats the simplified value of \(\large \dfrac{6}{2}\) ?

Answer 4

o why did you tell me before that it would be 3/1

Answer 5

can you confirm for the rest of the graphs

Answer 6

6/2 and 3/1 are same :

\(\large \dfrac{6}{2} = \dfrac{3\times 2}{2} = \dfrac{3}{1} = 3\)

Answer 7

and for b 1/8 =1/8+1.5

Answer 8

for c -4/1 = -4x+3

Answer 9

for d -1/8 = -1/8x + 1.5

Answer 10

wait, what exactly do you mean by : `b 1/8 =1/8+1.5` ?

The statements as-it-is makes no sense. If you want the equation of line, then you need to use slope intercept form :

\(\large y = mx + c\)

Answer 11

i mean 1/8x+1.5

Answer 12

For graph b, slope, \(m = \dfrac{1}{8}\)
y intercept, \(c = 1.5\)

So the equation of line becomes : \(\large y = \dfrac{1}{8}x + 1.5\)

Answer 13

can you help with integrals

Answer 14

Okay I'll try, ask...

Answer 15

http://prntscr.com/3s3i2r

Answer 16

can you show each integrals solution systematically

Answer 17

use below formula :

\[\large \int x^n ~dx = \dfrac{x^{n+1}}{n+1} \]


(*removed +C as we're dealing with definite integrals)

Answer 18

So start by changing the given fraction into above form ^

Answer 19

\[\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \]

Answer 20

see if that looks okay so far

Answer 21

yep

Answer 22

good, can you apply the earlier formula ?

Answer 23

\[\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \\ \large = \dfrac{x^{-2+1}}{-2+1} \]

Answer 24

are you still with me ? :)

Answer 25

yep

Answer 26

good, i forgot to put the bounds... let me fix it :

\[\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \\ \large = \dfrac{x^{-2+1}}{-2+1} \color{red}{\Bigg|_{-1}^1}\]

Answer 27

we need to take the difference of values at 1 and -1

Answer 28

ok

Answer 29

lets simplify the expression a bit before taking the difference

Answer 30

can you simplify the fraction ?

Answer 31

x^-1/-1

Answer 32

Excellent !
it simplifies even further..

Answer 33

-x^-1

Answer 34

good, i forgot to put the bounds... let me fix it :

\[\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \\ \large = \dfrac{x^{-2+1}}{-2+1} \color{red}{\Bigg|_{-1}^1} = \dfrac{x^{-1}}{-1} \color{red}{\Bigg|_{-1}^1} \]

Answer 35

it simplifies even further..

Answer 36

use below property of fractions :
\(\large a^{-n} = \dfrac{1}{a^n}\)

Answer 37

x^-1(-1)

Answer 38

good, i forgot to put the bounds... let me fix it :

\[\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \\ \large = \dfrac{x^{-2+1}}{-2+1} \color{red}{\Bigg|_{-1}^1} = \dfrac{x^{-1}}{-1} \color{red}{\Bigg|_{-1}^1} = -\dfrac{1}{x} \color{red}{\Bigg|_{-1}^1} \]

Answer 39

Now we're ready to take the difference :

when x = 1, whats the value of 1/x ?
when x = -1, whats the value of 1/x ?

Answer 40

x=1
x=1/-1

Answer 41

yes !

when x = 1, whats the value of 1/x = 1
when x = -1, whats the value of 1/x = -1

plug them and you're done !

Answer 42

\[
\large \int \limits_{-1}^1 \dfrac{dx}{x^2} = \int \limits_{-1}^1 \dfrac{1}{x^2} ~ dx= \int \limits_{-1}^1 x^{-2} ~ dx \\ \large = \dfrac{x^{-2+1}}{-2+1} \color{red}{\Bigg|_{-1}^1} = \dfrac{x^{-1}}{-1} \color{red}{\Bigg|_{-1}^1} = -\dfrac{1}{x} \color{red}{\Bigg|_{-1}^1} = -\left(1 -(-1)\right)
\]

Answer 43

simplify ^

Answer 44

-2

Answer 45

Correct !

Answer 46

next one

Answer 47

we only have to find for the upper bound right?

Answer 48

always find both bounds, the value becomes 0 at lower bound... so yes you don't need to find the lower bound for this particular problem. BUT the lower bound may not give 0 in other problems - So to be safe, always find the values at both bounds and take the difference okay ?

Answer 49

\[\large \int \dfrac{dx}{x^{\frac{1}{3}}}\]

Answer 50

step1 : change the fraction to \(\large x^n\) form

Answer 51

give it a try...

Answer 52

lower bound =0

Answer 53

upper bound ?

Answer 54

Ahh missed the bounds, one sec..

Answer 55

can you tell me for graph c how did we get -4

Answer 56

\[\large \int \limits_{0}^{\Pi}\dfrac{dx}{x^{\frac{1}{3}}}\]

Answer 57

for graph c ?

Answer 58

yes

Answer 59

Are you talking about graph c in that pic ?

Answer 60

tell me this :
rise = ?
run = ?

Answer 61

rise -4
run +1

Answer 62

why -4

Answer 63

Look at the graph -
Notice that running 1 unit to the right makes the line line to `sink` 4 units down

Answer 64

ok

Answer 65

graph a there also sink

Answer 66

You need to look at the graph : `left-to-right`

Answer 67

thats the key thing ^

Answer 68

reason is this - the values of number increase on a number line as you walk `left-to-right`

Answer 69

ok