Answer:
1- 16x-12
2- 6(6x-1)
3. -qx+24x-18
Step-by-step explanation:
Hope this help
Have a nice day
Step-by-step explanation:
f(x) = x² + x + 3/4
in general, such a quadratic function is defined as
f(x) = a×x² + b×x + c
the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 1
c = 3/4
x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =
= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =
= (-1 ± sqrt(2)i)/2
x1 = (-1 + sqrt(2)i) / 2
x2 = (-1 - sqrt(2)i) / 2
remember, i = sqrt(-1)
f(x) has no 0 results for x = real numbers.
for the solution we need to use imaginary numbers.
Answer:
483 with a remainder of 3
Step-by-step explanation:
4 8 3
---------------------------------
6 5 | 3 1 3 9 8
2 6 0
------------------
5 3 9
5 2 0
---------------------
1 9 8
1 9 5
------------------------------
3
Hope I explain myself :/
Answer:
y-5=11
y-5+5=11+5 (Add 5 to both sides)
y=16 (Simplify)
I believe that y=16 cannot be simplified so y=16 should be the answer let me know if it helped.
Hello,
Use the factoration
a^2 - b^2 = (a - b)(a + b)
Then,
x^2 - 81 = x^2 - 9^2
x^2 - 9^2 = ( x - 9).(x + 9)
Then,
Lim (x^2- 81) /(x+9)
= Lim (x -9)(x+9)/(x+9)
Simplity x + 9
Lim (x -9)
Now replace x = -9
Lim ( -9 -9)
Lim -18 = -18
_______________
The second method without using factorization would be to calculate the limit by the hospital rule.
Lim f(x)/g(x) = lim f(x)'/g(x)'
Where,
f(x)' and g(x)' are the derivates.
Let f(x) = x^2 -81
f(x)' = 2x + 0
f(x)' = 2x
Let g(x) = x +9
g(x)' = 1 + 0
g(x)' = 1
Then the Lim stay:
Lim (x^2 -81)/(x+9) = Lim 2x /1
Now replace x = -9
Lim 2×-9 = Lim -18
= -18
