Answer:
I Believe that your answer is D. Sorry if i'm wrong.
Step-by-step explanation:
Answer:
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
Start on the left side.
1
+
sec
2
(
x
)
sin
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
+
1
cos
2
(
x
)
sin
2
(
x
)
Write
sin
2
(
x
)
as a fraction with denominator
1
.
1
+
1
cos
2
(
x
)
⋅
sin
2
(
x
)
1
Combine.
1
+
1
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
sin
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
cos
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
Apply Pythagorean identity in reverse.
1
+
1
−
cos
2
(
x
)
cos
2
(
x
)
Simplify.
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1
cos
2
(
x
)
Now consider the right side of the equation.
sec
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
2
cos
2
(
x
)
One to any power is one.
1
cos
2
(
x
)
Because the two sides have been shown to be equivalent, the equation is an identity.
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
is an identity
Step-by-step explanation:
Answer:
Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.
Step-by-step explanation:
Remember that:
- Two lines are parallel if their slopes are equivalent.
- Two lines are perpendicular if their slopes are negative reciprocals of each other.
- And two lines are neither if neither of the two cases above apply.
So, let's find the slope of each equation.
The first basketball is modeled by:

We can convert this into slope-intercept form. Subtract 3<em>x</em> from both sides:

And divide both sides by four:

So, the slope of the first basketball is -3/4.
The second basketball is modeled by:

Again, let's convert this into slope-intercept form. Add 6<em>x</em> to both sides:

And divide both sides by negative eight:

So, the slope of the second basketball is also -3/4.
Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172