Answer:
Step-by-step explanation:
The probability is
P
(
A
)
=
1
3
Explanation:
To calculate the probability you have to count the number of all possible results
|
Ω
|
and the number of results that fulfill the given condition
|
A
|
.
In this case
Ω
=
{
1
,
2
,
3
,
4
,
5
,
6
}
Thereare 6 possible result of a dice toss. So
|
Ω
|
=
6
The given condition is "the result is divisible by
3
", so
A
=
{
3
,
6
}
- those are the only numbers divisible by
3
, so
|
A
|
=
2
.
Finally to calculate the probability we have to divide
|
A
|
by
|
Ω
|
.
P
(
A
)
=
|
A
|
|
Ω
|
=
2
6
=
1
3
Note The probability is never larger than
1
, so if you get such result then there must be a mistake in calculations.
Answer:
Substitute -8 as x into the equation.
h(-8)=-2(-8+5)^2+4
h(-8)=-2(-3)^2+4
h(-8)=-2(9)+4
h(-8)=-18+4
h(-8)=-14
:)
Answer:
Step-by-step explanation:
2log(x-3)=log25
log(x-3)²=log25 cancel out log on both sides
(x-3)²=25
x²-6x+9=25
x²-6x+9-25=0
x²-6x-16=0 factorize
x²+2x-8x-16=0
x(x+2)-8(x+2)
(x-8)(x+2)=0
x-8=0 or x+2=0
x=8 or x=-2
x=-2 cannot be considered because of the -ve sign.
x=8.
check
2log(x-3)
2log(8-3)
2log5
log5²=25
therefore LHS=RHS
Answer:
Check below, please.
Step-by-step explanation:
Hi, there!
Since we can describe eccentricity as 
a) Eccentricity close to 0
An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

b) Eccentricity =5

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

c) Eccentricity close to 1
In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

