Answer:
Step-by-step explanation:
As per the graph, the y-intercept is 20 which is the initial amount.
Another marked point has coordinates (4, 50).
<u>The slope is:</u>
<u>The equation of the line is:</u>
$20 is the initial deposit and weekly deposit is $7.5
Answer:


Step-by-step explanation:
Given

Solving (a): Write as inverse function

Represent a(d) as y

Swap positions of d and y

Make y the subject


Replace y with a'(d)

Prove that a(d) and a'(d) are inverse functions
and 
To do this, we prove that:

Solving for 

Substitute
for d in 




Solving for: 

Substitute 5d - 3 for d in 

Add fractions



Hence:

Answer:
k = 79
Step-by-step explanation:
3(k-99) = -60
(k-99) = -20
k= -20+99
k=79
Answer:
<em>132</em>
<em />
Explanation:
All you need to do is separately find the area of the shapes that are part of the figure (in this case the trapezoid and rectangle), and then add them all up.
1.(7x4) + (1/2x8[7+19])
2. 28 + 104
3. 132
~Hope this helps you~