Answer:
cdFGafadjgfsgdfzgFdSANJGHFHGFZSDFERQYWS6546QStep-by-step explanation:
Answer:
Step-by-step explanation:
I'm seeing that f(x) is
What you do to find an inverse is switch the x and y coordinates and then solve for the new y. Switching the x and y gives us:
Now we have to solve for the new y. Begin by muliplying both sides by 6 to get:
6x = 3y - 2 and
6x + 2 = 3y so
2x + 2/3 = y
To put it back into function notation:
Answer: 122?
Step-by-step explanation: i used a protractor (im not the best at that though but im 89% sure i'll be honest
Answer:
Our system of equations is:
We are looking for x
Let's express y using x
Replace x in the second equation with the result
- 4y-4x²-12x = -7
- 4(-2x-1)-4x²-12x = -7
- -8x-4-4x²-12x = -7
- -8x-4x²-12x = -7+4
- -4x²-20x = -3
- -4x²-20x+3 = 0 multiply by -1 to get rid of the - signs with x
- 4x²+20x-3=0
4x²+20x+3=0 is a quadratic equation
Let Δ be our discriminant
Δ= 20²-4*4*(-3)
Δ=448 > 0 so we have two solutions for x
let x and x' be the solutions
- x = = -5.145 ≈ -5.15
- x'= = 0.145≈ 0.15
so the solutions are:
-5.15 and 0.15
Answer: 3.751*10^18kg
Step-by-step explanation:
δ =619.09−0.000097p....equa1 where p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter.
Calculating the density of air at 5km above earth surface
P = 5000m + 6370000m = 6.375*10^6m
δ = 619.09 -(.000097* 6.375*10^6)
δ = 0.715kg/m^3 = density
Since Mass = density*volume...equ2
To calculate volume of air around the spherical earth at height 5km
V = (4/3 pai R^3) - (4/3pai r^3) ...equation 3 where R =6.375*10^6m, r = 6.37*10^6
Substituting R and r in equation 2 to solve for volume of air
V = 1.085*10^21 - 1.08*10^21
V = 5.25*10^18m^3
Substituting δ and V into equation 2 to solve for mass of air
M = 0.715 * (5.25*10^18)
M = 3.751*10^18kg