a) Was there ever a time when the first function = the second function?
Was there ever a time when 16x/x^2+2 = 12x/x^2+1 ?
16x / x2 + 2 = 12x / x2 + 1 I am guessing that this is supposed to be...
16x / (x2 + 2) = 12x / (x2 + 1) Cross-multiply.
16x(x2 + 1) = 12x(x2 + 2) Distribute.
16x3 + 16x = 12x3 + 24x Subtract 12x3 from both sides.
4x3 + 16x = 24x Subtract 24x from both sides.
4x3 - 8x = 0 Factor 4x out of both terms.
4x(x2 - 2) = 0 Set each factor equal to zero and solve for x .
4x = 0 or x2 - 2 = 0
x = 0 or x2 = 2
x = ± √2
Yes, they are the same when x = 0 , x = √2 , and x = -√2 .
b) For what values of x is 16x/x^2+2 greater than 12x/x^2+1 ?
16x / (x2 + 2) > 12x / (x2 + 1)
Look at the graph here.
We can see 16x / (x2 + 2) is greater than 12x / (x2 + 1) for x > √2 and for -√2 < x < 0 .
You might just need to say positive possibilities: x > √2
1) First we need to find the slope of l1 . Let's get its equation into slope-intercept form.
5x + 8y = -9 Subtract 5x from both sides of the equation.
8y = -5x - 9 Divide through by 8 .
y = -\(\frac58\)x - \(\frac98\)
Now we can see that the slope of l1 = -\(\frac58\) .
Line l2 is perpendicular to l1 , so the slope of l2 = +\(\frac85\) .
Line l2 has a slope of \(\frac85\) and passes through the point (10, 10) .
So.... in point - slope form, the equation of l2 is
y - 10 = \(\frac85\)(x - 10) We want this in the form y = mx + b . Distribute the \(\frac85\) .
y - 10 = \(\frac85\)x - \(\frac85\)(10)
y - 10 = \(\frac85\)x - 16 Add 10 to both sides.
y = \(\frac85\)x - 6
Now the equation for l2 is in the form y = mx + b , where m = \(\frac85\) and b = -6 .
We can look at a graph to verify that l1 and l2 are perpendicular, and l2 passes through (10, 10) .
m + b = \(\frac85\) + -6 = -4.4
A fish has a head 9 inches long. | → | h = 9 |
The tail equals the size of the head plus one-half the size of the body. | → | t = h + b/2 |
The body is the size of the head plus the tail. | → | b = h + t |
b = h + t Since h = 9 , we can substitute 9 in for h .
b = 9 + t
t = h + b/2 Substitute 9 + t in for b and 9 in for h .
t = 9 + (9 + t)/2 Multiply through by 2 .
2t = 18 + 9 + t
2t = 27 + t Subtract t from both sides.
t = 27
b = h + t Substitute 9 in for h and 27 in for t .
b = 9 + 27
b = 36
And the total length of the fish = h + b + t
h + b + t = 9 + 36 + 27 = 72 inches
I also get 72 inches, like AT and helperid.