Given:
bisects ∠RST.
To find:
The .
Solution:
Since, bisects ∠RST, therefore
...(1)
Now,
[Using (1)]
Therefore, the value of is
.
A - 6 - 6a = 4 find the value of a
2 answers:
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Answer:
The equation that matches the function shown is option;
C.
Step-by-step explanation:
The given graph of the function is a sinusoidal graph
The values of 'x' and 'y' coordinates at the maximum, x-intercept and minimum points are given as follows;
x, y
0 0
π 1
2·π 0
3·π -1
4·π 0
We note that sin(π/2) = 1, sin(π) = 0 sin(3·π/2) = -1, and sin(4·π/2) = sin(2·π) = 0
Therefore;
y = the sine of half the x-value
Which is presented as follows;
.
Answer:
cos(θ) = 3/5
Step-by-step explanation:
We can think of this situation as a triangle rectangle (you can see it in the image below).
Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.
Here we want to find cos(θ).
You should remember:
cos(θ) = (adjacent cathetus)/(hypotenuse)
We already know that the adjacent cathetus is equal to 3.
And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:
3^2 + 4^2 = H^2
We can solve this for H, to get:
H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5
The hypotenuse is 5 units long.
Then we have:
cos(θ) = (adjacent cathetus)/(hypotenuse)
cos(θ) = 3/5
If you are asking what is the graph of y = 3x^2 -2x+1.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.