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.
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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.
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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:
I believe it would be 4/6
Step-by-step explanation:
You have four different ways to get a sum of 7: 1 and 6; 2 and 5; 3 and 4; or 4 and 3.
Answer:

Step-by-step explanation:
Given Equation:
Equation:1
Equation:2
Dividing Equation:2 by '3' both the sides:
or
Equation:3
Putting the vale of 'x' in Equation:1


Subtracting '3' both sides



Putting value of 'y' in Equation:3


The solution of the equations is :

Let's find the least possibilities:
First number: 6
Second number: 8
Third number: 10
6 + 8 + 10 = 24
Now we can see that we are still missing 31 - 24 = 7.
7 can be gained by adding 3 and 4, so:
First number: 6 + 3 = 9
Second number: 8 + 4 = 12
Third number: 10
9 + 12 + 10 = 31
The linear function g which has a rate of change of -16 and initial value 600 is; g = -16x + 600.
<h3>What is the linear function which is as described?</h3>
It follows from the task content that the linear expression which is as described by the given verbal description is to be determined.
Since the standard slope-intercept form of a linear function is as given; f(x) = ax + b.
Where, a = slope (rate of change) and b = y-intercept (initial value).
Therefore, the required linear function which is as described is;
g = -16x + 600
Read more on linear functions;
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