<em>Greetings from Brasil...</em>
In a trigonometric function
F(X) = ±UD ± A.COS(Px + LR)
UD - move the graph to Up or Down (+ = up | - = down)
A - amplitude
P - period (period = 2π/P)
LR - move the graph to Left or Right (+ = left | - = right)
So:
A) F(X) = COS(X + 1)
standard cosine graph with 1 unit shift to the left
B) F(X) = COS(X) - 1 = -1 + COS(X)
standard cosine graph with 1 unit down
C) F(X) = COS(X - 1)
standard cosine graph with shift 1 unit to the right
D) F(X) = SEN(X - 1)
standard Sine graph with shift 1 unit to the right
Observing the graph we notice the sine function shifted 1 unit to the right, then
<h3>Option D</h3>
<em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>
Answer: 6.25 or 6 
Step-by-step explanation:
Lets do a pro gamer move and convert all fractions to decimals.
We now have 1+1+1.25+1.5+1.5.
From there on we simply do addition.
1+1+1.25+1.5+1.5=6.25
We get 6.25 or 6
Answer:
With probability you can multiply your chances in this case.
Getting an odd number: 1/2
Getting a number greater than 4: 1/6
Therefore it is a 1/12 chance to be both odd and greater than 4
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
hello :
<span>help :
<span>the discriminat of each quadratic equation :
ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 ) Δ > 0 the equation has two reals solutions : x =
(-b±√Δ)/2a</span>
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ < 0 : no reals solutions</span>
