Iwas wondering if anyone could me solve this undergraduate level optimization problem. prove that a

[tex]g(x)=\sqrt{2x-1}[/tex]
[tex]g(1)=\sqrt{2(1)-1}=\sqrt{2-1}=\sqrt{1}=1[/tex]
[tex]g(0)=\sqrt{2(0)-1}=\sqrt{0-1}=\sqrt{-1}=i[/tex]
[tex]g(5)=\sqrt{2(5)-1}=\sqrt{10-1}=\sqrt{9}=3[/tex]
[tex]g(-4)=\sqrt{2(-4)-1}=\sqrt{-8-1}=\sqrt{-7}=i\sqrt{7}[/tex]

[tex]f(x)=x^{2}-3x+1[/tex]
[tex]f(-3)=-3^{2}-3(-3)+1=9-(-9)+1=19[/tex]
[tex]f(0)=0^{2}-3(0)+1=1[/tex]
[tex]f(1)=1^{2}-3(1)+1=-1[/tex]

[tex]h(x)=3x-1[/tex]
[tex]h(2)=3(2)-1=5[/tex]

if you plug in 100 and 121 in the function, it equals 10 and 11.

Therefore your avereage rate of 1

Step-by-step explanation:

a)
3*1.99 I estimated 3*2=6
1*3.99=4
2*1.29=2*1.25=3
1.25*1
.20*3=.75
6+4+3+1+.75= 14.75
$15
I just used easy to add numbers that made sense
b)
3*1.99=4.97
+
1*3.99=3.99
+
2*1.29=2.58
+
1*1.25=1.25
+
0.20*3=0.60
=
13.39

Hey sorry I can't answer this but I just wanted to let you know that the butterfly died :( I couldn't pm because I have to answer more Qs but yea it died :(

I'm here to help.  Those 3 points all lie on the same line.  We know this because if we find the slope between any 2 of those points, it is the same.  I'll show you: [tex]m= \frac{12-20}{15-10}=- \frac{8}{5}[/tex].  Now another 2 points: [tex]m= \frac{4-12}{20-15}=- \frac{8}{5}[/tex].  The same holds for the first and third points.  So we got that out of the way.  Now we will pick any one of those points and use the x and y values and our slope to write an equation of that line and then finally solve for the x and y intercepts.  I'm going to use the first point (10, 20): [tex]y-20=- \frac{8}{5}(x-10)[/tex].  We will simplify to get it into y = mx + b form: [tex]y-20=- \frac{8}{5}x+ \frac{80}{5}[/tex] and [tex]y-20=- \frac{8}{5}x+16[/tex] and finally, [tex]y=- \frac{8}{5}x+36[/tex].  The y-intercept exists when x = 0, so when x = 0, y = 36.  The x-intercept exists when y = 0, so when y = 0, x = 22.5.  In summary, y-intercept: (0, 36).  x-intercept: (22.5, 0)  and you're all done!