Remark
The point value is (-2,5) So we know the two sides. We need the hypotenuse. We should notice that the x value is minus (-2) and value is y value is plus (5). That means we are in quad 2. Be careful how you read that. (-2,5) is a point. It is not a tangent.
Step One
Find the hypotenuse.
a = - 2
b = 5
c = ??
c^2 = a^2 + b^2
c^2 = (-2)^2 + 5^2
c^2 = 4 + 25
c^2 = 29 Take the square root of both sides.
sqrt(c^2) = sqrt(29)
c = sqrt(29)
Step Two
Find the Cosine of the angle.
Cosine(theta) = adjacent / hypotenuse
Cosine(theta) = -2 / sqrt(29) <<<<<<< Answer
Again, watch out for what you are given.
First of all we will eliminate x from our equations. In order to do that we will use our first and second equation and then we will use second and third equation.
Upon subtracting 2 equation from 1 we will get,
Now we will use second and third equation to eliminate x.
Adding 2nd and 3rd equation we will get,
Now we will find out y from our 4th and 5th equation.
Upon subtracting 5th equation from 4th equation we will get,
Now let us find out z by substituting y's value in 5th equation.
Now we will find x from by substituting y and z's value in equation 1.
Therefore, x=1, y=1 and z=0 is the solution of the given system.
A: 8 x 1/4 = 2, so Roger ran 2 miles. 5280 x 2 = 10560, so Roger ran 10560 feet.
B. Roger already ran 2 miles, so he needs to run 8 more miles to reach his goal. 1 lap = 1/4 mile, so divide 1/4 from 8 to get 32. Roger needs to run 32 more laps this week.
Answer:
y = -3/1x + 2
Step-by-step explanation:
1. Find the slope
The points I will use are (-1,5) and (0,2), where x1 = -1, y1 = 5, x2 =0, and y2 =2.
y2-y1/x2-x1
2-5/0-(-1) = -3/1
2. Find the y-intercepy
y = mx + b
You can use any x or y point. I'll use (-1,5)
m= slope, which is -3/1
5= -3/1(-1) +b
5= 3 + b
Subtract 3 from 5 and 3 to get 2=b.
3. Rewrite in slope-intercept form
y = -3/1x + 2
Silver for x
-3 - (-8) - (-2) = x
Switch the equation to make x on the right
x = -3 - (-8) - (-2)
x = 5 - (-2)
x = 7
• A negative minus a negative will always evaluate to a positive.
• A positive minus a negative will always evaluate to a positive.
