I am not completely sure on these
1.B
2.B
3.C
Answer:
Note from question, Let K be any integer. Integer = 1
θ = πk
θ = 3.142 * 1
θ = 3.142 in three decimal places
θ = sin⁻¹ (2/3) + 2kπ
θ = sin⁻¹0.667 + 2*1*3.142
θ = 0.718 + 6.284
θ = 7.002 in three decimal places
∴ 7.002 , 3.142
Step-by-step explanation:
Considering the equation
3 tan(θ) sin(θ) − 2 tan(θ) = 0
The objective is to solve the equation.
First solve the equation in one period.
3 tan(θ) sin(θ) − 2 tan(θ) = 0
( 3sinθ − 2 ) tanθ = 0
Therefore, 3sinθ − 2 = 0 also tanθ = 0
=> sinθ = 2/3 , tanθ = 0
Pick the right equation.
tanθ = 0
θ = tan⁻¹ 0
θ = 0
Using the unit circle, the period of tangent functions is π
Then the general solution of the equation is θ = 0 + πk ==> θ = πk
Pick the left equation.
3sinθ − 2 = 0
3sinθ = 2
sinθ = 2/3
θ = sin⁻¹ (2/3)
As the sine function has period 2π
Then the general solution is θ = sin⁻¹ (2/3) + 2kπ
Answer:
i) Line
ii) point
Step-by-step explanation:
Given is a number 3. We have to find the kind or kinds of symmetry
On seeing 3, we find that no vertical line can divide 3 into exactly two symmetrical figures.
But if we draw a horizontal line in the middle of 3, we get two symmetrical figures up and down the line.
Hence i) symmetric about a line
Next is imaging the middle of 3 end point.
At that point we find that the figure is symmetrical on all sides of the point
Hence there is also a point symmetry
Kinds of symmetry are
i) Line
ii) point
Answer:4.20 5.false
Step-by-step explanation:
Answer:
4.The answer is yes but not all of them are right triangles
can't seem to figure out five but anyway hope this helps
Step-by-step explanation: