Question

Help with Application question

Answer 1

Its Euler WARK!!!!!!!

Answer 2

^_^ Kupo

the graph becomes steeper since 0.045 > 0.039, this would be horizontal compression.

the second part can be answered logically (from real life experience) or by plugging in any value for v.
the smaller d(v) = smaller breaking distance for given v

Answer 3

so That means it would be compressed if it is steeper if i am correct

Answer 4

yes

Answer 5

Horizontally compressed

I think anyway because its .0039v which stands for f(x)= (ax) which i believe it means horizontally

Answer 6

actually since it is from 0.049 -> 0.039 it is vertical compression and not horizontal

0.039 is a fraction of 0.049

Answer 7

my bad for saying horizontal earlier

Answer 8

Dang it so close >.<

Answer 9

i forgot which was which. i used the bottom of this page and i think it's vertical compression now

http://www.regentsprep.org/regents/math/algtrig/ATP9/funclesson1.htm

Answer 10

I see, So about the 2nd part wouldn't it be less

Answer 11

f(x) --> f(ax) : horizontal change
f(x) --> af(x) : vertical change

Answer 12

horizontal:
a > 1 : compression
0 < a < 1 : stretch

vertical:
a > 1 : stretch
0 < a < 1 : compression

Answer 13

._. Wait so Its the opposite when its vertical compression

Answer 14

Alright that makes sense

Answer 15

Wark! So question, What would you do again to find the second part?

Answer 16

yes

Answer 17

you can, for this particular question, use logic: newer tires are more efficient than worn out tires. new tires require less breaking distance.

or you plug any value of v and see which has the smallest d(v)

plugging 1 should work and tell you right away that 0.045 > 0.039

so the distance for d(v) = 0.045v^2 will always be more than for d(v) = 0.039v^2

Answer 18

So one more so I don't get this wrong its D right XD in simpler terms

Answer 19

yes D

Answer 20

Thank you Moogle xD

Answer 21

Do you have time for more?

Answer 22

yes ^_^

Answer 23

:D WARK!!!!!

Answer 24

Would it be 5 + sqrt of 29

Answer 25

+or -

Answer 26

yes it is

Answer 27

:D

Answer 28

WARK!!!!!!!!!!!

Answer 29

alright as we move on

Answer 30

-12, 1?

Answer 31

yes

Answer 32

B

Answer 33

so B?

Answer 34

yup

Answer 35

just wondering. did you complete the square for the last 2 or deduced the right answer?

Answer 36

both could be done without any work :P

Answer 37

Yipeeeee

Answer 38

Yes I got that one answered, Is that the one that I called for you this morning?

Answer 39

did you try this one?
\[p(x) = ax^2 + bx + c\]
zeros are obtained with quadratic formula:
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 40

i don't know, i didn't see it

Answer 41

also this one would it D right?

Answer 42

the site was laggy so i couldn't follow the link and i lost it

Answer 43

i could find it again though

Answer 44

xD yeah this morning it was a mess

Answer 45

yes, D

Answer 46

YAY

Answer 47

x+6=25

Answer 48

\[a^2 + b^2 = c^2\]\[b = a + 6\]\[a^2 + (a+6)^2 = 25^2\]
you need to expand and solve for a

Answer 49

actually x + 36 = 625

Answer 50

x= 589

Answer 51

should i solve it or are you working on it?

Answer 52

brb 2 minutes

Answer 53

ehh i am confused now ._.

Answer 54

O WAIT YOU HAVE TO SQUARE IT

Answer 55

yes\[a^2 + (a+6)^2 = 25^2\]\[a^2 + (a^2 + 12a + 36) = 625\]\[2a^2 + 12a = 589\]

Answer 56

Oh boy

Answer 57

you're now looking at a quadratic

Answer 58

you can solve for a

Answer 59

Yes I believe I am

Answer 60

would you like to save your game?

Answer 61

2a^2=-12a +589

Answer 62

a^2= -6a +294.5

Answer 63

o_o??

Answer 64

you can do that

Answer 65

i think I screw that up

Answer 66

nope its fine

Answer 67

use the quadratic equation or complete the square to solve for a

Answer 68

It seems complicated to do that way, and I don't know where to go from there :/

Answer 69

from\[2a^2 + 16a = 589\]you can re-arrange it to\[2a^2 + 16a - 589 = 0\]the solutions to the quadratic equation, the roots, are the values of a (or x usually) when the polynomial = 0

for\[p(x) = ax^2 + bx + c = 0\]\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

our case a = 2, b = 12, c = -589. solve for x (which is really a, but not the same a, eh)

Answer 70

Sorry Openstudy is being weird again

Answer 71

wow its 2a^2 + 12a = 589 my bad

Answer 72

last sentence is still right though. only first 2 equations are wrong

Answer 73

that last part of the sentence o_O

Answer 74

lol :p

Answer 75

The a is not that a but another a o_o its A-ception

Answer 76

lol ya. i should have used capital letters for a b and c

Answer 77

let meh try to solve it then :p

Answer 78

cool :)

Answer 79

you'll always get 2 answers. sometimes only 1 makes sense based on context

Answer 80

(2 answers because of plus minus sign. 1 for each)

Answer 81

12 + or - sqrt of 4568 / 4 o_o???

Answer 82

[-12 (plus/minus) sqrt(4856)]/4

Answer 83

I forgot the negative *facepalm*

Answer 84

and got a dyslexic square root :P

Answer 85

lol

Answer 86

-12 + 68= 56

Answer 87

so a =14

Answer 88

-12 - 68= -80

Answer 89

a= -20

Answer 90

so A

Answer 91

yes