Rewrite the right side as √3/3 = 1/√3, and recall that tan(x) = 1/√3 when x = π/6. Then since tan is π-periodic, taking the inverse tan of both sides gives
where n is any integer. Solving for x, we get
and the solutions in the interval [0, 2π] are x = 5π/6 and x = 11π/6 (for n = 1 and n = 2).
The simple/ <span>common sense method:
</span>The typical lay out of a quadratic equation is ax^2+bx+c
'c' represents where the line crosses the 'y' axis.
The equation is only translated in the 'y' (upwards/downwards) direction, therefore only the 'c' component of the equation is going to change.
A translation upwards of 10 units means that the line will cross the 'y' axis 10 places higher.
9+10=19,
therefore <u>c=19</u>.
The new equation is: <u>y=x^2+19 </u>
<span>
<span>The most complicated/thorough method:
</span></span>This is useful for when the graph is translated both along the 'y' axis and 'x' axis.
ax^2+bx+c
a=1, b=0, c=9
Find the vertex (the highest of lowest point) of f(x).
Use the -b/2a formula to find the 'x' coordinate of your vertex..
x= -0/2*1, your x coordinate is therefore 0.
substitute your x coordinate into your equation to find your y coordinate..
y= 0^2+0+9
y=9.
Your coordinates of your vertex f(x) are therefore <u>(0,9) </u>
The translation of upward 10 units means that the y coordinate of the vertex will increase by 10. The coordinates of the vertex g(x) are therefore:
<u>(0, 19) </u>
substitute your vertex's y coordinate into f(x)
19=x^2+c
19=0+c
c=19
therefore <u>g(x)=x^2+19</u>
A:
Area: 289 m
Perimeter: 51 m
B:
Area: 4 m
Perimeter: 8 m
A) The given equation has no solution. The absolute value cannot be negative, but must be -9 in order for the equation to be satisfied.
b) |x -7| = 2 . . . . . . . the equation
Solution 1:
-2 = x -7
5 = x . . . . . . add 7
Solution 2
x -7 = 2
x = 9 . . . . . . add 7
The two numbers are {5, 9}
Answer:
it is a mixed number the number is 1 1/2
Step-by-step explanation:
