Help with Pre-Calculus problems. Can anyone help
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Help with Pre-Calculus problems. Can anyone help

[From: ] [author: ] [Date: 11-08-23] [Hit: ]
............
Just about finished with 100 problem review packet but still need help with a few. Would like it if I could receive answer with all of the steps/work. Want to see how exactly answer is reached. I will award necessary points if needed or multiple if needed (will try to see how to do that as well XD). Please help

Simplify.

e^( 2+ln x )

Value of

(arc cos(1/2))

Solve for x, where x is the real number

log(x) + log( x - 9 ) = 1

.................1- 2x, x<1
Let f(x)= { Find f(4) and find f(-5)
................-x^2, x ≥ 1

Identify the x & y intercepts and identify the domain and range. (no work required for these)

y = x^2 - 6x + 8

y = ³√x

y = |x + 3| - 2


........x^2 if x < 0
y = { x + 2 if 0 ≤ x ≤ 4
........5 if x > 4

Please need to finish this by weds. Your saving a life ;)

Note just all the periods you see to push part of the problem so it would diagonally fit with rest of the problem

-
1) e^(2+lnx)=e^2* e^lnx= e^2 *x= xe^2

2) arccos(1/2)= "the angle between 0 and pi, with cos = 1/2"
= pi/3 or 60 degrees

3)logx + log(x-9)=1...using properties of logs
Log(x(x-9))= 1
10^1= x(x-9)

X^2-9x-10=0. Now factor and check.

(x-10)(x+1)=0
x=10 or -1
-1 is not in the domain of logx , so x= 10

4) this is a piece function, where the domain is divided.
Since 4>1, f(4)=-4^2= -16 ( not (-4)^2)

Since -5 <1, f(-5)= 1-2(-5)= 11

5)y=x^2-6x+8
Domain: all reals, range : y>= -1
X intercepts (2,0) and (4,0)
Y intercept (0,8)

6) y= x^(1/3). Domain and range: all reals
Xand y intercept: (0,0)

7)y=|x-3|+2

Domain : all reals; range y>= 2
X intercept (none)
Y intercept (0, 5)

8) this is also a piece function with three pieces. Are you supposed to graph?
If so , draw the parabola y = x^2, but only the left side. Start at (0,0) with an open circle. Connect it to (-1,1), (-2,4), etc to infinity.
The middle piece is a segment of a line. It starts with a closed circle at (0,2) and ends with a closed circle at (4,6)
The last piece is part of a horizontal line at y =5. It starts with an open circle at 4,5) passing through (5,5), (6,5) , etc to infinity.

If you need to find values, like f(3), just see which piece of the function it belongs to.
f(3)= 3+2=5

Maybe you wanted domain, etc? Oh well.
Domain : all reals ; range: y>0
x intercepts: none; y intercept (0,2)

Good luck!

Hoping this helps!
1
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