The number beside the letter . for example 6x, 6 would be the variable
<span>Simplifying
(6a + -8b)(6a + 8b) = 0
Multiply (6a + -8b) * (6a + 8b)
(6a * (6a + 8b) + -8b * (6a + 8b)) = 0
((6a * 6a + 8b * 6a) + -8b * (6a + 8b)) = 0
Reorder the terms:
((48ab + 36a2) + -8b * (6a + 8b)) = 0
((48ab + 36a2) + -8b * (6a + 8b)) = 0
(48ab + 36a2 + (6a * -8b + 8b * -8b)) = 0
(48ab + 36a2 + (-48ab + -64b2)) = 0
Reorder the terms:
(48ab + -48ab + 36a2 + -64b2) = 0
Combine like terms: 48ab + -48ab = 0
(0 + 36a2 + -64b2) = 0
(36a2 + -64b2) = 0
Solving
36a2 + -64b2 = 0
Solving for variable 'a'.
Move all terms containing a to the left, all other terms to the right.
Add '64b2' to each side of the equation.
36a2 + -64b2 + 64b2 = 0 + 64b2
Combine like terms: -64b2 + 64b2 = 0
36a2 + 0 = 0 + 64b2
36a2 = 0 + 64b2
Remove the zero:
36a2 = 64b2
Divide each side by '36'.
a2 = 1.777777778b2
Simplifying
a2 = 1.777777778b2
Take the square root of each side:
a = {-1.333333333b, 1.333333333b}</span>
Answer:
a = 12.93
Step-by-step explanation:
Convert to decimal: 12 1/3 = 12.3
Pythagoras (a^2 = b^2 + c^2):
4^2 = 16
12.3^2 = 151.29
16+151.29 = 167.29
a = √167.29
a = 12.93 (rounded to two decimal places)
<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>
