Exercise 1:
exponential decay:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 600
We look for b:
(480/600) * (100) = 80%
b = 0.8
Substituting:
y = 600 * (0.8) ^ ((1/3) * t)
We check for t = 6
y = 600 * (0.8) ^ ((1/3) * 6)
y = 384
Answer:
exponential decay:
y = 600 * (0.8) ^ ((1/3) * t)
Exercise 2:
linear:
The function is given by:
y = ax + b
Where,
a = -60 / 2 = -30
b = 400
Substituting we have:
y = -30 * x + 400
We check for x = 4
y = -30 * 4 + 400
y = 280
Answer:
linear:
y = -30 * x + 400
Exercise 3:
exponential growth:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 512
We look for b:
(768/512) * (100) = 150%
b = 1.5
Substituting:
y = 512 * (1.5) ^ ((1/2) * t)
We check for t = 4
y = 512 * (1.5) ^ ((1/2) * 4)
y = 1152
Answer:
exponential growth:
y = 512 * (1.5) ^ ((1/2) * t)
Answer:
35 < 56.
- <em>56 is greater than 35.</em>
<u><em>I hope this helped at all.</em></u>
Answer:
the calibration of the system
Step-by-step explanation:
7 mobiles
Answer:
The standard form of a hyperbola with vertices and foci on the x-axis:

where:
- center: (h, k)
- vertices: (h+a, k) and (h-a,k)
- Foci: (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
- Slopes of asymptotes:

<h3><u>Part 1</u></h3>
The center of the given hyperbola is (0, 0), therefore:

Therefore
are the vertices. From inspection of the graph,
.
<h3><u>Part 2</u></h3>
Choose two points on the asymptote with the positive slope:
(0, 0) and (4, 6)
Use the slope formula to find the slope:

<h3><u>Part 3</u></h3>
Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

Therefore, 
<h3><u>
Part 4</u></h3>
Substitute the found values of
and
into the equation from part 1:


48 is the answer. Hope I'm the brainliest!