


→ Substitute cos 2x and sin 2x by their expressions

→ Subtract both sides by sin x

→ Take sin x as a common factor in the right side

From the graph, the equation has 4 solutions, the intersection points between the 2 graphs
Answer:
Part 1) The length of the diagonal of the outside square is 9.9 units
Part 2) The length of the diagonal of the inside square is 7.1 units
Step-by-step explanation:
step 1
Find the length of the outside square
Let
x -----> the length of the outside square
c ----> the length of the inside square
we know that

step 2
Find the length of the inside square
Applying the Pythagoras Theorem

substitute



step 3
Find the length of the diagonal of the outside square
To find the diagonal Apply the Pythagoras Theorem
Let
D -----> the length of the diagonal of the outside square




step 4
Find the length of the diagonal of the inside square
To find the diagonal Apply the Pythagoras Theorem
Let
d -----> the length of the diagonal of the inside square




Answer:Two such terms are 7x^3*y^9 and -3x*y^5
Their quotient is
7x^3*y^9
--------------
-3x*y^5
This can be simplified as follows:
The numerical coefficients become -7/3.
x^3/x = x^3*x^1 = x^(3 - 1) = x^2 (we subtract the exponent of x in the denominator from the exponent of x in the numerator).
Next, y^9*y^5 = y^4.
The quotient in final reduced form is then (-7/3)x^2*y^4
Answer: 100
Step-by-step explanation:
4b squared where b= 5
Slot in 5 for b; 5 squared = 5x5= 25
4 (25)
=100
Answer:
the equation would be: y=(x-4)(x+2)