Question

need help on part 2 review!!!!

Answer 1

you can't assume everyone knows what you are talking about.

Answer 2

Go ahead and post your question and I will try to explain.

Answer 3

i know iwas about to upload da file thwen my CPU froze

Answer 4

Okay. All of it or do you have something specific?

Answer 5

all of it please

Answer 6

Okay. Well the first problem with the table. For that one, it is simply plug and chug. So since it gives you a bunch of x and y values. You can take your answer choice and plug in the x values and see if it gives you the corresponding y values on the table.

Answer 7

For the next you, when you set the equations equal you want to know where they intersect. So if it asks for the x value, simply find the x that falls directly UNDER the intersection. In this case: 2

Answer 8

For the next one, the range is the y values. So, simple list the y values of the points: {2,4,5,6} 5 appears twice but you only need to list it once.

Answer 9

soc on number 3 is wrong right

Answer 10

No, c is correct.

Answer 11

To test for symmetry about the origin. You need to see if for a point (a,b) there exists another point (-a,-b) So for this problem, plug in 1 for example, then -1. You get: (-1,2) but when you plug in positive 1 you get. (1,-2). So it is symmetric about the origin.

Answer 12

ok i understand that and did i get number one correct???

Answer 13

Yup :)

Answer 14

okok and on #2 x = -2

Answer 15

No, on this one x=2. Because if you go to the x directly under the intersection of the two lines (not the axes!) x is +2.

Answer 16

ok thank u so much u wanna help me with the other page??

Answer 17

Sure, as long as you click good answer for me :P haha

Answer 18

You can just upload the page if thats easiest :)

Answer 19

cool i can do dat ur a big help i have to get dis packet done or else and its up loading now

Answer 20

alright, no problem at all :)

Answer 21

oki think this is it

Answer 22

thats the same page

Answer 23

no wrong one hold on

Answer 24

k here we go

Answer 25

Okay. For the first one, I'm just going to go down the answer choices:
For for A: Its zero is 3.
Zeros are where the graph crosses the x-axis. (also known as x-intercepts). But, according to this graph you can see that it does NOT cross at 3. This is incorrect.
For B: It's minimum value is -6. Here, you want to see where the lowest Y value is. And if you look all the way at the bottom, the lowest point IS -6.
For C: On (1,6] the function is decreasing (i.e., negative slope).
For D: The domain is all the x values for the graph. As you can see, x ranges from -4 to 6. So the WRONG one (which is NOT correct) would be answer choice A. Since zeros refers to the x intercepts NOT the y intercepts.

Answer 26

For the second one, you notice that f(x) would be 3x? Take any x value and multiply it by 3 and you get the corresponding y value. So, it is ONLY positive integers (and zero). So this makes the answer choice A because it contains ONLY integers greater than or equal to zero.

Answer 27

For the 3rd one, you are correct. The rate is NOT constant. (Tell me if you need an explanation)

Answer 28

Then for the last one, a geometric sequence is something in the form. ar^n So in this case you have: 10,40,160,640. Factor out a 10: 10(1,4,16,64...) And you see you have 10(4^n) Which is what is underlined.

Answer 29

Does that help?

Answer 30

yea gima sec im workin it out

Answer 31

Okay :P

Answer 32

ok ok thank u ready for the next?? ive got 7 pages to do want 2 continu??

Answer 33

Haha, I'll do one more then I have to go :P

Answer 34

cool ur a big help im uploadin it now

Answer 35

Thank you :)

Answer 36

here we r

Answer 37

Alright. For the first one, you can factor it or use the quadratic formula. I notice that -10 factors to -5 and +2. So: (x-5)(x+2). That means that your solutions are x=5 and x=-2 because (5-5)=0 and (-2+2)=0. You just have your signs switched.

Answer 38

For the next one, the solutions are wherever Y is zero. So according to the chart that is x=5 and x=7. Same as the previous problem, you're setting y=0.

Answer 39

For the next one. Start my multiplying both sides. So:
\[(4+k)(6)=(k-2)(8)\]. Distribute.
\[24+6k=8k-16\].
Add the 16 and subtract the 6k:
\[40=2k\].
Divide by 2.
k=20. Which you have.

Answer 40

For the next one, square both sides:
\[(\sqrt{3z+1})^2=5^2\].
\[3z+1=25\].
Subtract the 1
3z=24
Divide by 3.
z=8. Which you also have.

Answer 41

And for the last one, multiply out the left side:
\[(x+3)(x+3)=x^2+6x+9=2\].
Subtract the 2.
\[x^2+6x+7=0\].
This time, you have to use the quadratic formula.
\[\frac{-6 \pm \sqrt{6^2-4(1)(7)}}{2(1)}\].
This gives:\[-3 \pm \frac{\sqrt{8}}{2}\].
But \[\sqrt{8}=2\sqrt{2}\].
\[-3 \pm \frac{2\sqrt{2}}{2}=-3 \pm \sqrt{2}\].
Which you have.

Answer 42

ok thanks HI FIVE ur awesome man!!!!! thank u soo much i feel retricethought i might pass math now ur awesome!!!! dontlet anyone tell u otherwise

Answer 43

Haha, thank you :P I'm in my senior year of college as a math/physics major :P Sorry for the lengthy explanations. I hate just giving away answers I prefer understanding :P Good luck man! I'm sure you'll do fine!

Answer 44

thanks gud luck in college u will pass it for me i gotta make it 2 10th grade