Question

Help pleaseee MEDAL! <3

Answer 1

k

Answer 2

ok

Answer 3

whats up?

Answer 4

1. solve the system of equations.
y=2x^2-3
y=3x-1 (1 point)
A. no solution
B. (-1/2,5)(2,-5/2)
C. (-1/2,-5/2)(2,5)
D. (1/2,5/2)(2,5)

Answer 5

@Marco,Phillip

Answer 6

2. How many real number solutions are there to the equation 0=-3x^2+x-4
A. 0
B. 1
C. 2
D. 3

Answer 7

it'd be better if you post only one question at a time :)

Answer 8

3. Solve the equation by completing the square. If necessary, round to the nearest hundredth. x^2-18x=19
A. 1;19
B.-1;19
C.3;6
D.-3;1

Answer 9

lets start with 1.
y=2x^2-3
y= 3x-1
so,

2x^2-3 = 3x-1
this is a simple quadratic equation.
do you know how to solve quadratics ?

Answer 10

ohh okay, ill just wait until those get answered to put the last few :) thanks for all of the help and please dont just put the letter id really love it if you could explain to me how to get the answer, this stuff confuses me >.<

Answer 11

Sorta

Answer 12

so can you try to solve
2x^2-3 = 3x-1
?
try it, if you get stuck, i am here :)

Answer 13

so either im really confused or the answer is no solution for #1, anyone know if im correct?

Answer 14

and number 2 is 2 solutions? anyone out there to tell me if im getting these right?

Answer 15

let me give you a headstart,
2x^2 -3x-2=0

2 numbers with sum =-3 and product=-4 are .???

sorry, I keep on losing the internet connection :(

Answer 16

\(2x^2 -3x-2=0 \\ 2x^2-4x+x-2=0 \\ 2x(x-2) +1(x-2) = 0\)

can you try to continue?
factor out x-2 from on the left side.