If we know that f(-4) = 8, then what (x, y) point must lie on the graph y = f(x)?.
The point is (-4, 8)
For a given function y = f(x), all the points (x, y) on the graph are of the form:
(x, y) = (x, f(x))
So if we know that:
f(-4) = 8
Then we know that the point:
(-4, f(-4)) = (-4, 8)
Must lie on the graph of y = f(x)
Then the correct answer is (-4, 8)
If you want to learn more, you can read:
Write definition for each one and draw the triangle .
1) Right-angled triangle:
2) Acute:
3) Obtuse:
Answer:
right angle triangle 90 degrees
scute less than 90 degrees
obtuse over 90 degrees
Step-by-step explanation:
Here are the drawings
y=5x
solve by substitution
x-2y=-2
y=2x+4
solve by elimination
x+2y=-7
x-5y=7
PLEASE HELP WORTH 10 POINTS! AKA SHOW WORK:)
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
0,5,-3,-7