thelizzybeth

So let x equal of one of the acute angles in a 90 degree triangle. Since \(tan=\frac {opposite}{adjacent}= \frac {5}{1}\), and by the Pythagorean Identity, we can find that the sides are 5,1, and \(\sqrt 26\). Therefore, cos x = \(\frac{adjacent}{hypotenuse}\)= \(\frac{1}{\sqrt 26}\) = \(\frac{\sqrt 26}{26}\).

The shortest distance is the edge of a cube plus a diagonal of the base square. (Draw a diagram, and after a bit of trial and error & intuition, you should see why). So the answer is \(1+\sqrt2\). 

\(\frac{AD*BC}2=\frac{BE*AC}2\) because in a triangle, all of the altitudes * the corresponding bases divided by two equal the area. They are in the same triangle, so the areas must be equal. 

Now, substitute the values and solve for BE. 

You should be able to finish the arithmetic. 

Let's start off by finding the equation of the line perpendicular to \(y=\frac{1}{2}x+5\). To do this, we note that the slope of two perpendicular lines is the negative reciprocal of each other. So we have \(y=-2x+b\). Now we plug in the point (6,2) to find that b=14. Our perpendicular line is \(y=-2x+14\). 

Next, we find the intersection point of the two lines by solving for the two equations. We get the point \((\frac{18}{5}, \frac{34}{5})\). 

Now, we can use the Mid-Point Formula to find our desired point, since \((\frac{18}{5}, \frac{34}{5})\) must be the midpoint of our desired point (a, b) and (6,2).

\(\frac{a+6}{2}=\frac{18}{5}\) and \(\frac{b+2}{2}=\frac{34}{5}\). 

Solve for these two equations and you have your point (a, b)!

2. We start off by rotating the point (-3, 2) 90 degrees clockwise around the origin. Now the coordinates are (2,3). Reflecting this point over the line y=x means flipping the x and y values. Therefore, the coordinates of this new point C is (3,2).

1.

a) First, from the circle equation, we can see that the center of the circle is (2, -3). Now we use the distance from point to line formula to figure out the distance from the center of the circle to the line. 

Formula: the distance between (x₁, y₁) and a line of the form ax + by + c = 0 is

\(\frac{|ax_{1}+by_{1}+c|}{\sqrt(a^2+b^2)}\)

Now, plugging the values in, \(\frac{|-8-15+20|}{\sqrt(16+25)}\), and solving for this equals \(\frac{3}{\sqrt(41)}\). Rationalizing the denomiator, we get that the answer is \(\frac{3\sqrt(41)}{41}\).

b) The greatest distance is the distance we just found above plus the circle's radius. (To see why, draw a diagram- it's just a line from the end of the circle to the point)

However, your equation implies that the circle's radius is the square root of -4, which is not a real number. Did you type it wrong?

We're going to turn each sentence into an equation.

a = number of $1, b= number of $2, c= number of $3

b=a+11

c=a-18

a+2a+3b=100

Substituting the first two equations into the third equation,

a+2a+22+3a-54=100

Simplifying, we get 6a=132, a=22

So there are 22 one-dollar bills.

Alright, let's write a couple of equations.

First, let x equal the speed at which she went to the other town (what we're trying to find) and let y equal the time it takes her.

Then, because distance=speed*time, 150=xy

Our second equation is going to be on her way back. 150=(x+15)(y+\(\frac{1}{2}\)). We put \(\frac{1}{2}\)hr instead of 30 minutes because our speeds are given in km/hrs. 

Now we expand: 150=xy+\(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)

Substitute 150=xy (our first equation). 150=150+\(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)

This simplifies to \(\frac{1}{2}\)x+15y+\(\frac{15}{2}\)= 0

Multiply by two: x+30y+15=0

From our first equation, y=(150/x), and substituting this into our above equation:

x+30(150/x)+15= x+(4500/x)+15=0

Multiply by x: x²+15x+4500=0

Factorize to get (x+60)(x−75)=0

We get either -60 or 75. Speed must be positive, so the answer is 75. 

I'll try to get through as many problems as I can!

3. This is going to be a bit tricky to explain without a diagram, but just draw one yourself on a piece of paper and follow along.

Let's start off with finding the cylinder's height.

First, draw the cylinder's space diagonal. The space diagonal of the cylinder is going to be the circle's diameter, which equals 10. 

Now draw the diameter of the cylinder's base. The diameter of the base, the space diagonal, and the height form a right triangle. Using the Pythagorean Theorem, we find that the height equals the square root of 10²-6² , which is 8.

Now we can figure out the volume of the cylinder: 3²\(\pi\)*8= 72\(\pi\)

Now we find the sphere's volume: \(\frac{4}{3}\pi\) * 5³ 

The space inside the sphere and outside the cylinder equals the sphere's volume - the cylinder's volume, which you should be able to calculate.

Ahh! That's clever.