2)
P(4,-4) -->(-4, 7)
4 - 8 = -4 -------->left 8
-4 + 11 = 7 -------->up 11
Answer: left 8; up 11
3)
C(3,-1) , left 4 up 1
3 - 4 = -1 -------->left 4
-1 + 1 = 0 -------->up 1
a)
(x , y) -->(x - 4 , y +1)
C(3, -1) -->C'(-1 , 0)
b)
(x , y) --> (x - 4, y + 1); (-1 , 0)
If Keith is represented by x, and Joe is three less than Keith, Troy is represented by x-3.
Then, we know that Joe exercises for twice the amount of time as Troy, we multiply the expression for Joe by two.
Keith = x
Troy = x-3
Joe = 2(x-3)
Answer:
b=5√3, c=10
Step-by-step explanation:
30-60-90 triangle
c=2a
b=a√3
A word to the wise: It's <span> f(x)=125(0.9)^x, where ^ represents exponentiation.
In this case the ave. value over the interval [11, 15] is
125(0.9)^15 - 125(0.9)^11
------------------------------------- = (125/4) [ 0.9^15 - 0.9^11)
15 - 11 = (31.25) [ 0.2059 - 0.3138 ] = a negative result
= (31.25)(-0.1079) = -3.372 (av. r. of c.
over the interval [11,15] )
Do the same thing for the time interval [1,5]. Then compare the two rates of change.</span>
Hey I would first do the one in parentheses so 7-5 is 2 so your equation is now 60/5*2 so now what will you do?
