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If \(\frac{9x^3\:+\:4x^2\:+\:11x\:+\:7}{x^2\:+\:bx\:+\:18\:}\) is real, its denominator must never equal zero

The discriminant of x^2+bx+18 is b^2-72

if x^2+bx+18 doesn't equal zero, than: 

b^2-72<0

b=+/-(6 sqrt (2))

so the greatest integer is 8

JP

thanks for the correction

first write \(y=\frac{3x+20}{4}\) in standard form: 

\(y=\frac{3x+20}{4}\:\: \\4y=3x+20\:\:\:\:\:\:\:\:\:\: \\3x-4y+20=0\:\:\)

using distance from point to line formula: 

\(\frac{\left|3\cdot 0-4\cdot 0+20\right|}{\sqrt{3^2+\left(-4\right)^2}}=\frac{20}{5}=4\)

call this distance RS

rs is a perpendicular bisector of the chord

this creates a right triangle with sides AR, RS, and AS

so the radius squared  is: \(10^2+4^2=116\)

area of circle: \(\pi r^2=116\pi \)

JP

the only way the expression \(\frac{x^2\:+\:1}{x^2\:-\:4x\:+\:c}\) is not real is if the denominator equals zero

 \(x^2\:-\:4x\:+\:c=0 \)

discriminant is:

 \((-4)^2-4c \\=16-4c\)

to have no real solutions:

\(16-4c<0 \\4c>16 \\c>4\)

smallest integer value is c=5

JP

You can use a calculator and get about 193.994845224, which when rounded to the nearest integer is is 194

equation of the red line in the form y=mx+b: 

slope is 2, y-intercept is -1, so the equation of the red line is y=2x-1

since it's a dashed line, the line isn't included in the shaded region

so the equation is y<2x-1

in the form ax+by+c>0, this equation sould be 2x-y-1>0

JP

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{18} \\x+y=\frac{xy}{18} \\18x+18y=xy \\xy-18x-18y=0 \\xy-18x-18y+324=324 \\(x-18)(y-18)=324\)

factors of 324: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, and 324

can you do it yourself now? 

JP

Remember, distance=rate*time

                          Distance                 Rate              Time

First part of trip        30                       9                  30/9

Last Part of trip        10                       x                  10/x

Whole trip                40                       10                  4

(Time for the first part of the trip) + (time for the second part of the trip) = (total time of trip)

\(\frac{30}{9}+\frac{10}{x}=4 \\30x+90=36x \\-6x=-90 \\x=15\)  

so the cyclist's rate must be 15 kph in the next 10 km in order to bring the overall speed up to 10 km per hour

JP

Can you explain how you got your answer?

if x

if x>=a, the graph y=f(x) is the same graph as y=ax^2

line in the graph has positive slope, since the parobola only has nonegative values

so, a>0, and thus, the line in the graph passes through every single horizantal line <= a^2+2a

the parabola region passes through every single horizantal line >=a^3

a^2+2a>=a^3

a+2>=a^2 (since a>0 and can divide by a)

0>a^2-a-2

0>(a-2)(a+1)

-1<=a<=2

greatest value of a is 2

JP