LO-U10T2 – #2: I can graph a cosine and secant function using transformations.
a)
Write the equation in standard form
f(x) = a cos[b(x - h)] + k where b 2 0 or f(x) = a sec[b(x - h)] + k where b > 0.
b) State the base equation and transformations.
c)
Create a table of points that moves the base graph through each of its transformations. You
must have 5 points in your table.
d) Sketch the graph using the points found in c). Be sure that the scale and spacing on the axes is
appropriate. Make your graph the size of at least 4 of a page.
e)
State the period, phase shift, and amplitude.
f) State the Domain and Range.
COSX
1. a)
-X +
cos(-x) =
c)
b) basery cost
2005[x - 5] + 3
horiz shift
pase V+E ky
'vert shrink 12
10,0
reflect across x-axis
(
10)
II, OJ
right It
(F/2, 1) 17/21/2/17 - 12/17 U
ITO (TO JT, 3/2]
(81, 1/2 / 131, 121 122 z
(21/312)
ادور 2ار

$450 is the answer.

step-by-step explanation:

30% of 450 = 135

450-135=315

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."