1.) Find an exponential function of the form y=ab^x whose graph passes through the points (2,48) and (5,3072)
2.) The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x=3.
x=6, y=-1
3.) The variable z varies jointly with x and y. Use the given values to write an equation relating x, y, and z. Then find z when x =-2 and y=4.
x=1/3, y=3/4, z=5/2
1.) To find the exponential function of the form y = ab^x that passes through the points (2, 48) and (5, 3072), we need to find the values of a and b.
We can start by substituting the first point (2, 48) into the equation:
48 = ab^2
Next, substitute the second point (5, 3072) into the equation:
3072 = ab^5
Now we have a system of two equations. We can solve for a and b simultaneously using the method of substitution or elimination.
From the first equation, we can isolate a:
48 = ab^2
a = 48/b^2
Substitute this expression for a in the second equation:
3072 = (48/b^2)b^5
3072 = 48b^3
b^3 = 64
Taking the cube root of both sides, we find:
b = 4
Now substitute this value of b back into the first equation:
48 = a(4^2)
48 = a(16)
a = 48/16
a = 3
Therefore, the exponential function that passes through the points (2, 48) and (5, 3072) is:
y = 3 * 4^x
2.) If x and y vary inversely, it means that their product remains constant. Let's write an equation:
x * y = k
Substitute the given values x = 6, y = -1, and solve for k:
6 * (-1) = k
-6 = k
The equation relating x and y is:
x * y = -6
To find y when x = 3, substitute x = 3 into the equation and solve for y:
3 * y = -6
y = -2
Therefore, when x = 3, y = -2.
3.) If the variable z varies jointly with x and y, it means that z is directly proportional to both x and y. Let's write an equation:
z = k * x * y
Substitute the given values x = 1/3, y = 3/4, and z = 5/2, and solve for k:
5/2 = k * (1/3) * (3/4)
5/2 = k * 1/4
5/2 = k/4
k = 10
The equation relating x, y, and z is:
z = 10 * x * y
To find z when x = -2 and y = 4, substitute these values into the equation and solve for z:
z = 10 * (-2) * 4
z = -80
Therefore, when x = -2 and y = 4, z = -80.
1.) To find an exponential function of the form y=ab^x that passes through the points (2,48) and (5,3072), we can use the equation y=ab^x and substitute the coordinates into the equation to form a system of equations.
First, substitute the coordinates (2,48):
48 = ab^2
Second, substitute the coordinates (5,3072):
3072 = ab^5
Now we have a system of equations with two unknowns (a and b). We can solve the system to find the values of a and b.
Divide the second equation by the first equation to eliminate a:
3072/48 = (ab^5)/(ab^2)
64 = b^3
Take the cube root of both sides to solve for b:
b = 4
Substitute this value of b into the first equation to find a:
48 = a(4^2)
48 = 16a
a = 3
So the exponential function of the form y=ab^x that passes through the points (2,48) and (5,3072) is y = 3 * 4^x.
2.) If x and y vary inversely, we can write an equation relating x and y as xy = k, where k is a constant.
Using the given values x=6 and y=-1, we can substitute them into the equation to find k:
(6)(-1) = k
-6 = k
So the equation relating x and y is xy = -6.
To find y when x = 3, we can substitute x=3 into the equation and solve for y:
(3)y = -6
3y = -6
y = -2
Therefore, when x=3, y=-2.
3.) If z varies jointly with x and y, we can write an equation relating x, y, and z as z = kxy, where k is a constant.
Using the given values x=1/3, y=3/4, and z=5/2, we can substitute them into the equation to find k:
5/2 = k(1/3)(3/4)
5/2 = k(1/4)
k = (5/2)/(1/4)
k = 10
So the equation relating x, y, and z is z = 10xy.
To find the value of z when x = -2 and y = 4, we can substitute these values into the equation:
z = 10(-2)(4)
z = -80
Therefore, when x = -2 and y = 4, z = -80.
#1
since y=ab^x, just plug in the two points you are given:
ab^2 = 48
ab^5 = 3072
Divide the 2nd by the first and you have
b^3 = 64
b = 4
a*4^2 = 48, so a = 3 and
y = 3*4^x
#2 inverse variation means xy = k, a constant. Plug in your values and you have
(6)(-1) = (3)(y)
y = -2
#3 direct variation means z = kxy
Or, z/xy = k, a constant. Plug in your values and you have
(5/2)/((1/3)(3/4)) = z/((-2)(4))
z = 80