<span>1) Name the variables
Number of days: x
rent: y
2) state the initial points
x y
days $
3 285
60 510
3) assume linear relation:
=> (y - yo) / (x - xo) = (y1 - yo) / (x1 - xo)
=> (y - 285) / (x - 3) = (510 - 285) / (60 - 3)
=> (y - 285) / (x - 3) = 225 / 57 = 75 / 19
=> 19 (y - 285) = 75 (x - 3)
=> 19y - 19*285 = 75x - 75*3
=> 19y - 75x = 5415 - 225
=> 19y - 75x = 5190
=> standar form = -75x + 19y = 5190
PART B: Write the equation obtained in Part A using function notation.
-75x + 19y = 5190
=> 19y = 5190 + 75x
=> y = 5190/19 + (75/19)x
=> function notation = f(x) = (75/19)x + 5190 / 19
PART C: Describe the steps to graph the equation obtained above on the
coordinate axes. Mention the labels on the axes and the intervals.
1) Coordinate axes:
x: number of days
y: rent
2) draw the two given points: (3,285) and (60, 510)
3) draw the line that joins those points from the interception of the y-axis until some points further (60, 510).
</span>
A car travels - 1/6.
in 3/5 of an hour -
60 / 5 - 12 x 3 = 36.
so in 36 minutes.
4/5 = 36+12 = 48 minutes.
2/6.
5/5 (full hour ) = 60 minutes.
3/6.
Car can travel - 3/6.
Answer:
+
*LN(|
|) +C
Step-by-step explanation:
we will have to do a trig sub for this
use x=a*tanθ for sqrt(x^2 +a^2) where a=2
x=2tanθ, dx= 2 sec^2 (θ) dθ
this turns 
into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ
the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1
then it simplifies into integral(4*sec^3 (θ)) dθ
you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C
then you will need to rework your functions of θ back into functions of x
tanθ will resolve back into 
(see substitutions) while secθ will resolve into 
sec(θ)= is from its ratio identity of hyp/adj where the hyp. is 
and adj is 2 (see tan(θ) ratio)
after resolving back into functions of x, substitute ratios for trig functions:
= + *LN(||) +C
Answer:
Step-by-step explanation:
Given
and 
Required
The midpoint (M)
This is calculated as:
So, we have:
Answer:
The mean absolute value deviation is a measure of dispersion which gives the average variation of the data from the mean. In order words, it is the average of the positive distances of each point from the mean. The larger the Mean Absolute Deviation, the greater variability there is in the data (the data is more spread out).
The MAD helps determine whether the set's mean is a useful indicator of the values within the set.
The larger the MAD, the less relevant is the mean as an indicator of the values within the set.
Step-by-step explanation:
