12/5 is sequences that’s the answer
<span>1) Number of fish in the pond: y
Number of years: x
</span>A pond had an initial population of 120 fish. Then when x=0, y=120. The graph must begin at point (x,y)=(0,120). Only the graph above at right and below at left begin in this point (0,120).
<span>The number of fish is exponentially decreasing by one-fourth each year. Then the first year the number of fish must decrease 120*(1/4)=120/4=30, and the number of fish after the first year must be:
</span>120-30=90=120(1-1/4)=120(4-1)/4=120(3/4). Then when x=1, y=90. The point is (x,y)=(1,90)
In the graph above at right when x=1, y is between 24 and 36. y=90 is not in this interval. then this graph is not the correct.
In the graph below at left when x=1, y is between 84 and 96. y=90 is in this inverval. Then this is the correct graph.
Answer: T<span>he graph of the solution set for this situation is the graph below at left.
2) The equation has the form:
y=y0(r)^x
Where y0 is the initial population and r is the rate of reduction. In this case:
y0=120 and
r=1-1/4=(4-1)/4→r=3/4
Then the equation modeled by the graph is:
y=120(3/4)^x
Answer: The equation modeled by the graph is that above at right:
y=120(3/4)^x</span>
Answer:
Taater is a good friend of mine.
Step-by-step explanation:
The answer would be the third option, hope I'm just as helpful in times of need if its possible to subtract those then it's 100% the first option. Sorry don't exactly know this, so I'm at least trying to help.
Answer:
lines 1 and 3
Step-by-step explanation:
y = 2 is a horizontal line parallel to the x- axis
x = - 4 is a vertical line parallel to the y- axis
Then these 2 lines are perpendicular to each other
y = 15x - 3 ( in the form y = mx + c ) with m = 15
y + 1 = - 5(x + 2) ( in the form y - b = m(x - a) with m = - 5
For the lines to be perpendicular the product of their slopes = - 1
However
15 × - 5 = - 75 ≠ - 1
The 2 lines 1 and 3 are perpendicular
