Answer:
5
Step-by-step explanation:
We know that 13 is the hypotenuse. With this we can use the Pythagorean Theorem to solve this problem.
Let the third side be x.
13^2=12^2+x^2
169=144+x^2
169-144=x^2
25=x^2
x=5
Solved.
We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
</span>- 1/(2-c)=-1/-(c-2)=1/(c-2)
(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2)
(c-4-1)/(c-2)=(c-2)/(c+2)---------------- > (c-5)/(c-2)=(c-2)/(c+2)
(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14
the solution is c=14
the domain of the function is (-∞,-2) U (-2,2) U (2,∞) or
<span>all real numbers except c=-2 and c=2</span>
First draw sloping up (steep because he’s running)to the right until you hit 20 on the y axis. Next draw down and to the right for 5 yards until 15 on y axis. Next is a flat horizontal line for 3 seconds on x axis. Finally a shallow slope line down (because he is walking) and right until you hit the x axis.
Answer:
-4/5
Step-by-step explanation:
To find the slope of the tangent to the equation at any point we must differentiate the equation.
x^3y+y^2-x^2=5
3x^2y+x^3y'+2yy'-2x=0
Gather terms with y' on one side and terms without on opposing side.
x^3y'+2yy'=2x-3x^2y
Factor left side
y'(x^3+2y)=2x-3x^2y
Divide both sides by (x^3+2y)
y'=(2x-3x^2y)/(x^3+2y)
y' is the slope any tangent to the given equation at point (x,y).
Plug in (2,1):
y'=(2(2)-3(2)^2(1))/((2)^3+2(1))
Simplify:
y'=(4-12)/(8+2)
y'=-8/10
y'=-4/5
