Y=mx+b
b=y intercept
the current one is y=-2
6 units up is -2+6=4
new one is 4
In the standard form of the equation
![\\ \ f(t)=Acos[b(t\pm c)]+k\\ \\](https://tex.z-dn.net/?f=%20%5C%5C%20%5C%20f%28t%29%3DAcos%5Bb%28t%5Cpm%20c%29%5D%2Bk%5C%5C%20%5C%5C%20)
The middle line =k
For our given problem
f(t) = 40cos (80t + 20)
On comparison we get k=0
Hence middle line=0
<h3><u>The value of the greater number is 15.</u></h3><h3><u>The value of the smaller number is 7.</u></h3>
x = 2y + 1
3x = 5y + 10
Because we have a value for x we can plug this value in to find the value of y.
3(2y + 1) = 5y + 10
Distributive property.
6y + 3 = 5y + 10
Subtract 5y from both sides.
y + 3 = 10
Subtract 3 from both sides.
y = 7
We can plug this value back into the original equation to find the value of x.
x = 2(7) + 1
x = 15
Distance between points

and

is

Distance from H to B:
![[tex]d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}](https://tex.z-dn.net/?f=%5Btex%5Dd%3D%5Csqrt%7B%2810-%28-3%29%29%5E2%2B%281-%28-9%29%29%5E2%7D%3D%5Csqrt%7B169%2B100%7D%3D%5Csqrt%7B269%7D)
d=\sqrt{(1-(-3))^2+(10-(-9))^2}=\sqrt{16+361}=\sqrt{376}[/tex] units.
Distance from Z to B:

units.
Horse Z is closer to the barn.
(The conversion to meters is not required; the question does not ask for actual distances, so "units" is OK.)