Answer:
Correct answer: C (6.5 , - 8)
Step-by-step explanation:
Given points A(-1,-2) and B(9, -10)
The formula by which we calculate the coordinates of a point, let be C, that divides the line segment in the given ratio is.
Xc = (Xa + λ · Xb) / (1 + λ) and Yc = (Ya + λ · Yb) / (1 + λ)
Where λ is the coefficient which depends on the direction of the ratio in this case is:
λ = 3 : 1 = 3
Xc = (-1 + 3 · 9) / (1 + 3) = (-1 + 27)/ 4 = 26 / 4 = 6.5
Xc = 6.5
Yc = (-2 + 3 · (-10) / (1 + 3) = (-2 - 30)/ 4 = -32 / 4 = - 8
Yc = - 8
God is with you!!!
Answer:
A. (3,1)
B. g(x)=|x-3|+6
C. h(x)=-|x-3|-6
Step-by-step explanation:
A. To graph the absolute value function f(x) = |x - 3| + 1, first graph the parent absolute value function y=|x| and then translate it 3 units to the right and 1 unit up (see green graph in attached diagram). The vertex of the function f(x) is at point (3,1).
B. The function g(x) translates f(x) 5 units up, so its equation is
g(x)=f(x)+5
g(x)=|x-3|+1+5
g(x)=|x-3|+6
Blue graph in attached diagram.
C. The function h(x) reflects g(x) over the x-axis, so the equation of the function h(x) is
h(x)=-g(x)
h(x)=-(|x-3|+6)
h(x)=-|x-3|-6
Red graph in attached diagram.
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
Answer:
when you subtract 123-42, you get 81
Step-by-step explanation: