Answer:
−6a − 35b + 48c
Step-by-step explanation:
You do pemdas and if you know Pemads you will get your anwser
Answer:
x = 2 cm
y = 2 cm
A(max) = 4 cm²
Step-by-step explanation: See Annex
The right isosceles triangle has two 45° angles and the right angle.
tan 45° = 1 = x / 4 - y or x = 4 - y y = 4 - x
A(r) = x* y
Area of the rectangle as a function of x
A(x) = x * ( 4 - x ) A(x) = 4*x - x²
Tacking derivatives on both sides of the equation:
A´(x) = 4 - 2*x A´(x) = 0 4 - 2*x = 0
2*x = 4
x = 2 cm
And y = 4 - 2 = 2 cm
The rectangle of maximum area result to be a square of side 2 cm
A(max) = 2*2 = 4 cm²
To find out if A(x) has a maximum in the point x = 2
We get the second derivative
A´´(x) = -2 A´´(x) < 0 then A(x) has a maximum at x = 2
Answer:
Statements 3, 4 and 5 are true.
Step-by-step explanation:
x^2 - 8x + 4
Using the quadratic formula:
x = [ -(-8) +/- √((-8)^2 - 4*1*4)] / 2
= (8 +/- √(64 - 16)) / 2
= 4 +/- √48 / 2
= 4 +/- 4√3/2
= 4 +/- 2√3.
So the third statement is true.
Converting to vertex form:
x^2 - 8x + 4
= (x - 4)^2 - 16 + 4
= (x - 4)^2 -12
So the extreme value is at (4, -12)
So the fourth statement is true.
The coefficient of the term in x^2 is 1 (positive) so the graph has a minimum.
First combine like terms ,
5/5x=9/5x
move the variable which will cancel out because it is the same
5=9 is no solution sorry if i’m wrong lol