Answer:
( f h ) (x) = 6 x² - 1
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given<em> f(x) = 3 x - 4</em>
g (x) = −x²+2 x−5
<em> h (x) = 2 x² + 1</em>
j (x) = 6 x + 2 - 8 x
K (x) = 3 x² - x + 7
<u><em>Step(ii)</em></u>:-
<em>( f h ) (x) = f ( h (x)) = f ( 2 x² + 1 )</em>
= 3 (2 x² + 1 ) - 4
= 3 ((2 x² ) + 3 - 4
= <em>6 x² - 1</em>
<u><em> Final answer:</em></u>-
∴ <em> ( f h ) (x) = 6 x² - 1</em>
F(x) = <span>x^2+3x+8
now, the pending of the tangent line is d/dx f(x)
f'(x) = 2x + 3
now, we need know when the pending is increasing.
so
</span>2x + 3> 0
solving
x>-3/2
The interval over which the function f(x)= x^2+3x+8 is <span>increasing is (-3/2,+</span>∞<span>)</span><span>
</span>
Answer:
(1, 5)
Step-by-step explanation:
The midpoint of a line segment has coordinates that are the average of the end-point coordinates.
M = (X + Y)/2 = ((-3, 4) +(5, 6))/2 = (-3+5, 4+6)/2 = (2, 10)/2
M = (1, 5)
The midpoint is (1, 5).
Hey there!
Do PEMDAS
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
15 + 2(2 * 4) - 4^2
= 15 + 2(8) - 4^2
= 15 + 16 - 4^2
= 31 - 4^2
= 31 - 4 * 4
= 31 - 16
= 15
Therefore, your answer is: 15
Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
Answer:
x² + 6x - 5x – 30=(x-5)(x+6)
Step-by-step explanation:
x² + 6x - 5x – 30=x²+x-30
delta=1-4(-30)=1+120=121=11²
Sqrt(delta)=11
The roots are:
x1,2=(-1 +- 11)/2
Thus we get
x1=(-1+11)/2=10/2=5
x2=(-1-11)/2=-12/2=-6
Finally:
x² + 6x - 5x – 30=(x-5)(x+6)
