Every time teams play with each other (assuming there are no draws) half of them win
and the others get eliminated so
64÷2=32 32 games played and 32 teams eliminated
32÷2=16 16 games played and 16 teams eliminated
16÷2=8 the same goes here
8÷2=4 and here
4÷2=2 .
2÷2=1 .
32+16+8+4+2+1=63 games
Hello :
the <span>coordinates </span> circumcenter for ∆DEF :
x = (1+7+1)/3
y = (1+1+5//3
x=3
y= 7/3
If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV. We need to focus on Q IV due to the restrictions on theta.
Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis. Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2
The secant of theta is the reciprocal of that, and thus is
2 sqrt(2)
---------- * ------------ = sqrt(2) (answer)
sqrt(2) sqrt(2)
Answer:
Step-by-step explanation:
Polynomial f(x) has the following conditions: zeros of -4 (multiplicity 3), 1 (multiplicity 1), and with f(0) = 320.
The first part zeros of -4 means (x+4) and multiplicity 3 means (x+4)^3.
The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).
The third part f(0) = 320 means substituting x=0 into (x+4)^3*(x-1)*k =320
(0+4)^3*(0-1)*k = 320
-64k = 320
k = -5
Combining all three conditions, f(x)
= -5(x+4)^3*(x-1)
= -5(x^3 + 3*4*x^2 + 3*4*4*x + 4^3)(x-1)
= -5(x^4 + 12x^3 + 48x^2 + 64x - x^3 - 12x^2 - 48x - 64)
= -5(x^4 + 11x^3 + 36x^2 + 16x -64)
= -5x^3 -55x^3 - 180x^2 - 80x + 320
Answer:
C) -1.255
Step-by-step explanation:
We are tasked to solve for the residual value given that when x equals 29, y will be equals to 27.255. But, when it is tested, y actual value is 26. The formula in solving residual is shown below:
Residual value = Observed value - predicted value
Residual value = 26 - 27.255
Residual values = -1.255
The answer is -1.255 for residual value.
