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)
9514 1404 393
Answer:
x = 12.75
∠C = 51.75°
Step-by-step explanation:
The two marked angles are complementary, so ...
(5x -12) +(3x) = 90
8x = 102
x = 102/8
x = 12.75
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C = (5x -12)° = (5(12.75) -12)°
∠C = 51.75°
__
<em>Check</em>
E = 3x = 38.25, then C+E = 51.75 +38.25 = 90 . . . . as it should
Https://us-static.z-dn.net/files/d4c/b1c4613c4e210c04f4bcda5810cab46d.png
Answer:
7
Step-by-step explanation:
add 7 to 3x
Then: 4x-3x=1x which gives:
1x=7
7/1=7
