Help with num 4 please.​

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)

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|