Okay so this is a very hard conceptual question. We need to prove that (x, y) is the ordered pair when "f(x) = g(x)".
"f(x) = g(x)" represents the point where the lines share a point or basically the intersection point of the two functions.
To prove that the intersection point is (x, y) let's find the x and y values at the point of intersection.
f(x) ----> the x-value is x and the y-value is f(x)
g(x) -----> the x-value is x and the y-value is g(x)
We know that f(x) = g(x) so we know that the y values match too.
We can also substitute a variable y for f(x) or g(x) (It is simply the y-value when x is plugged in for x. I know it sounds a bit confusing.).
So the solution when f(x) = g(x) is (x, y)!!!
Step-by-step explanation:
no of the above madam...
Answer:
125/6(In(x-25)) - 5/6(In(x+5))+C
Step-by-step explanation:
∫x2/x1−20x2−125dx
Should be
∫x²/(x²−20x−125)dx
First of all let's factorize the denominator.
x²−20x−125= x²+5x-25x-125
x²−20x−125= x(x+5) -25(x+5)
x²−20x−125= (x-25)(x+5)
∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx
x²/(x²−20x−125) =x²/((x-25)(x+5))
x²/((x-25)(x+5))= a/(x-25) +b/(x+5)
x²/= a(x+5) + b(x-25)
Let x=25
625 = a30
a= 625/30
a= 125/6
Let x= -5
25 = -30b
b= 25/-30
b= -5/6
x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)
∫x²/(x²−20x−125)dx
=∫125/6(x-25) -∫5/6(x+5) Dx
= 125/6(In(x-25)) - 5/6(In(x+5))+C
Greetings ! Your Answer Is Below.
Answer;
1. m = 16
2. x = -z + 34 or z = -x + 34
3. p = 49
Step-by-step explanation;
1. m + 13 - 13 = 29 - 13 | Subtract 13 from both sides ( m = 16 )
2. - z + 34 | Add -z to both of the sides ( x = - z + 34 )
Or - x + 34 | Add -x to both of the sides ( z = - x + 34 )
3. 136 - 87 = 49 | Subtract 87 from both sides ( p = 49 )
[ Hope This Helped ! ]
Answer:
21 + 12i
Step-by-step explanation:
Given
(3 - 4i)(1 + 5i) - (2 - i) expand the product of factors
= 3 + 11i - 20i² - 2 + i
= 3 + 11i + 20 - 2 + i collect like terms
= 21 + 12i
