W varies partly as x and partly as the square of x. When x=3, w=18 and when x= 5, w=169. Find w in terms of x.
w=ax+bx^2+c
169=5a+25b+c
18=3a+9b+c
So, assume c=0 first
find a and b.
multiply first equation by 3, second by 5
169*3=15a+75b
90=15a+45b
subtract second equation from first
507-90=30b
b=13.9
then find a in the second equation
then, go to the original statement, check it with your a and b.
I can't be able to answer it pls help me out
To find w in terms of x, we first need to determine the relationship between w and x.
Given that w varies partly as x and partly as the square of x, we can write the equation as:
w = kx + mx^2
where k and m are constants.
Next, we can substitute the given values to form a system of equations:
When x = 3, w = 18:
18 = 3k + 9m (equation 1)
When x = 5, w = 169:
169 = 5k + 25m (equation 2)
Now, we can solve this system of equations to find the values of k and m.
Multiply equation 1 by 5 and equation 2 by 3 to create similar terms:
90 = 15k + 45m (equation 3)
507 = 15k + 75m (equation 4)
Subtract equation 3 from equation 4 to eliminate k:
507 - 90 = 15k + 75m - 15k - 45m
417 = 30m
Divide both sides of the equation by 30:
417/30 = 30m/30
13.9 = m
Now that we know the value of m, we can substitute it back into equation 1 to solve for k:
18 = 3k + 9m
Substitute m = 13.9:
18 = 3k + 9(13.9)
18 = 3k + 125.1
Subtract 125.1 from both sides of the equation:
18 - 125.1 = 3k + 125.1 - 125.1
-107.1 = 3k
Divide both sides of the equation by 3:
-107.1/3 = 3k/3
-35.7 = k
We have now determined the values of k and m:
k = -35.7 and m = 13.9.
Finally, we can substitute these values back into the original equation to find w in terms of x:
w = kx + mx^2
w = (-35.7)x + (13.9)x^2
Therefore, w in terms of x is:
w = -35.7x + 13.9x^2
To find the relationship between w and x, we'll need to express w in terms of x.
Let's start by setting up the equation:
w = k * x^a * x^b
Where k is a constant, a represents how w varies with x, and b represents how w varies with the square of x.
Given that when x = 3, w = 18, we can substitute these values into the equation:
18 = k * 3^a * 3^b
Simplifying this equation further, we have:
18 = k * 3^(a + b)
Similarly, when x = 5, w = 169:
169 = k * 5^a * 5^b
Simplifying again:
169 = k * 5^(a + b)
Now we have a system of two equations:
18 = k * 3^(a + b)
169 = k * 5^(a + b)
To eliminate k, we can divide the second equation by the first:
169/18 = (k * 5^(a + b)) / (k * 3^(a + b))
9.39 = (5^(a + b)) / (3^(a + b))
Next, we can rewrite the equation in terms of the base:
(5/3)^(a + b) = 9.39
To isolate (a + b), we can take the logarithm of both sides:
log((5/3)^(a + b)) = log(9.39)
(a + b) * log(5/3) = log(9.39)
(a + b) = log(9.39) / log(5/3)
Finally, substitute back into one of the initial equations. Let's use the first equation with x = 3 and w = 18:
18 = k * 3^a * 3^b
18 = k * 3^(a + b)
Substituting the value of (a + b) we found above:
18 = k * 3^log(9.39) / log(5/3)
Now, solve for k:
k = 18 / (3^log(9.39) / log(5/3))
Finally, we have the value of k. Substituting this value back into the equation W = k * x^a * x^b, we can find w in terms of x.