Let the points be A,B,C
#A
(leg3)
(leg1)
(leg2)
#B
We have to find perimeter
#C
The length of leg3=17.4units
Answer:
y=1/2(x-1)
Step-by-step explanation:
If x=t^2 and t>0, then t=sqrt(x).
If t=sqrt(x) or x^(1/2) and y =1-1/t, then y=1-x^(-1/2).
The x-intercept is when y=0.
So we need to solve 0=1-x^(-1/2) to find point P.
Add x^(-1/2) on both sides: x^(-1/2)=1.
Raise both sides to -2 power: x=1
So point P is (1,0).
Let's find tangent line at point (1,0).
We will need the slope so let's differentiate.
y'=0+1/2x^(-3/2)
y'=1/(2x^(3/2))
The slope at x=1 is y'=1/(2[1]^(3/2))=1/(2×1)=1/2.
Recall point-slope form is y-y1=m(x-x1).
So our line we are looking for is y-0=1/2(x-1)
Let's simplify left hand side y=1/2(x-1)
T=10.8, was the multiplication Symbol a part of the quest
left means subtracting
6 7/8 - 2 5/8
Whole numbers subtract each other
Numerators subtract each other
The denominators stay the same.
6-2= 4
7-5=2
4 2/8
Reduce 2/8. Divide by 2. 2/2= 1, 8/2= 4
4 2/8=4 1/4
Answer: 4 2/8 = 4 1/4 cups left
Slope: 15 Y Int:40 equation y=15x+40. 24 Hours $55
