I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
SEE ATTACHED IMAGE.
Part A:
The differential equation is:
y '= 0.08y
The initial condition is:
P (0) = 500 Part B:
The formula for this case is:
P (t) = 500exp (0.08 * t) Part C:
After five days we have to evaluate t = 5 in the equation.
We have then:
P (5) = 500 * exp (0.08 * 5)
P (5) = 745.9123488
Rounding:
P (5) = 746
The solution is y = -5
What is Quadratic Equation?
A quadratic equation is a second-order polynomial equation in a single variable x , ax² + bx + c = 0. with a ≠ 0. Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has at least one solution. The solution may be real or complex.
Given data ,
Let the function be f ( x ) = y
And , y = x² + 2x - 8
So , f ( x ) = x² + 2x - 8
Substituting the value for x in the equation , we get
When x = -3
f ( -3 ) = ( -3 )² + 2( -3 ) - 8
= 9 - 6 - 8
= 9 - 14
f ( -3 ) = -5
When x = 1
f ( 1 ) = ( 1 )² + 2( 1 ) - 8
= 1 + 2 - 8
= 3 - 8
f ( 1 ) = -5
Hence , the value of y = -5
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Answer:
The solution of given equation are -1 and 5.
Step-by-step explanation:
The given equation is

We need to solve the above equation by finding the zeros of

The vertex form of an absolute function is

where, a is constant and (h,k) is vertex.
Here, h=2, k=-3. So vertex of the function is (2,-3).
The table of values is
x y
0 -1
2 -3
4 -1
Plot these points on a coordinate plane and draw a V-shaped curve with vertex at (2,-3).
From the given graph it is clear that the graph intersect x-axis at -1 and 5. So, zeroes of the function y=|x-2|-3 are -1 and 5.
Therefore the solution of given equation are -1 and 5.
Now solve the given equation algebraically.

Add 3 on both sides.


Add 2 on both sides.

and 
and 
Therefore the solution of given equation are -1 and 5.
Answer:
1x0
2 x 11
1x0
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