Range is output for given input
domain is D={-3,0,5}
input -3,0 and 5 for x
f(-3)=(-3)^2-6(-3)+4
f(-3)=9+18+4
f(-3)=31
F(0)=(0)^2-6(0)+4
f(0)=0-0+4
f(0)=4
f(5)=(5)^2-6(5)+4
f(5)=25-30+4
f(5)=-1
Domain={-3,0,5}
Range={31,4,-1}
first, find the numeric value for 11/15
second to find theta, simply do the <em>inverse</em> cos (which is cos^-1) of the first answer.
now you know theta, just do the sin of 90 - theta and that's it!
since you know whatr cos(theta) is, you just take the inverse cos of that number to get theta and 'reverse' cos, essentially. you are just solving for theta, by reversing the cos function with cos^-1
please mark as brainliest!
Step-by-step explanation:
The center of a circle with 2 end points of a di diameter is the midpoint of the two endpoints.
The formula needed to find the minpoints is
(x,y) = (x2 + x1)/2, (y2 + y1)/2
x2 = 3
x1 = 3
y2 = 0
y1 = -7
midpoint = (3 + 3)/2, (0 - 7)/2
midp[oint = 3,-3.5
The midpoint is the center of the circle. Observe that the signs get changed when entering the values for (x,y)
So far what you have is (x - 3)^2 + (y + 3.5)^2 = r^2
To determine r^2 you need only take the distance from the center to oneof the endpoints.
r^2 = (3 - 3)^2 + (3.5 - 0)^2
r^2 = 3.5^2
r^2 = 12.25
Answer: (x - 3)^2 + (y + 3.5)^2 = 12.25
X² + 10x +2 = 0 .Complete the square of (x² + 10x + ??)
We know that (10x) is equal to twice the square root of the 1st (x) by ??
10x = 2.x.(??) and ?? = 5 and its square is 25. Then I have to add 25 and subtract 25 so that not to change the equation:
(x² + 10x + 25) -25 +2 =0
(x+5)² -23
X^4 - 4x^3 - 9x^2 + 36x = 0
x(x^3 - 4x^2 - 9x + 36) = 0
x(x^2(x - 4) - 9(x - 4) = 0
x(x + 3)(x - 3)(x - 4) = 0
the zeroes are -3, 0, 3, 4 Answer