Question

help ill post attachments

Answer 1

that first one is uh.. an accident lol

Answer 2

@AZ

Answer 3

One question at a time, my dude. If you post them all at once, no one's going to bite since that's too daunting.

1)
We want to solve it algebraically. We have two equations that both start with y = something

we can use the substitution method

basically
if a = b
and a = c

then that means b = c

So we have
y = 3x + 8
y = 3x^2 + 14x - 12

can you set them equal to each other and solve the quadratic function?

Answer 4

no i can't i'm dumb

Answer 5

3x + 8 = 3x^2 + 14x - 12

can you bring 3x to the other side by subtracting it on both sides?

and do the same thing to the 8? Subtract 8 on both sides

so that way we can get one side to be equal to 0

Answer 6

no i literally can't i'm pretty dumb

Answer 7

lol

Answer 8

if all you can do is make excuses, you won't get very far

3x + 8 = 3x^2 + 14x - 12

let's subtract 8 on both sides first

3x + 8 - 8 = 3x^2 + 14x - 12 - 8

on the left side of the = sign, +8 - 8 = 0

we're left with

3x = 3x^2 + 14x - 12 - 8

can you simplify it?

what is -12 - 8 = ?

Answer 9

-4

Answer 10

wow az the math man :)

Answer 11

no

-12 MINUS 8

-12 + 8 would have been -4

-12 - 8 is going to be what?

Answer 12

oh lmao told you i'm dumb haha wow

Answer 13

if you take out the negative sign for a minute, it's the same thing as

-(12 + 8)

Answer 14

-20

Answer 15

Exactly, so now we have

3x = 3x^2 + 14x - 20

now let's subtract 3x on both sides

we're left with

0 = 3x^2 + 14x - 3x - 20

what is 14x - 3x = ?

Answer 16

11x

Answer 17

Good! So now we have

3x^2 + 11x - 20 = 0

To find out the solutions, we need to use the quadratic formula

\( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \)

when the equation is \( ax^2 + bx + c = 0\)

so for your question
a = 3
b = 11
c = -20

can you plug it in to the formula?

Answer 18

x =
-11± pi -11^2 -43-20

Answer 19

hmm

some formatting errors, and no idea where that 'pi' came from

\( \dfrac{-11\pm \sqrt{11^2 - 4(3)(-20)}}{2(3)}\)

can you simplify it?

what is
11^2
what is 4 * 3 * -20

what is 2 * 3

Answer 20

121
-240
6

Answer 21

So now we have

\(\dfrac{-11 \pm \sqrt{121 - (240)}}{6}\)

what is 121 - (-240) =

Answer 22

361

Answer 23

chris hemsworth is kinda hot

Answer 24

Now take the square root of 361

Answer 25

how u do that

Answer 26

19

Answer 27

google is my best friend

Answer 28

yes

so we have

\(\dfrac{-11 \pm 19}{6}\)

so the two x-values are

(-11+19)/6 ??

and

(-11-19)/6 = ??

Answer 29

1.33

Answer 30

or should i round it to 1.34

Answer 31

you can round to 1.34

so that's for (-11+19)/6

what about when it's (-11-19)/6

because remember we had a \(\pm\) so we have two answers, one where we add the two numbers and another where we're doing subtraction

Answer 32

-5

Answer 33

yes

so the two x-values are 4/3 (or 1.34) and -5

now we need to find the y-value for those x-values

just choose any equation, y = 3x + 8 is the easier one and plug in x = 4/3 and x = -5

what are the y-values?

Answer 34

idk.

Answer 35

i don't know how to plug in in math but i know how to plug in in bed

Answer 36

y = 3x + 8

plug in x = 4/3

y = (3*4/3) + 8

y = ??

Answer 37

12

Answer 38

so for x = 4/3 then y = 12

so one solution is (4/3 , 12)

now what about when x = -5

y = (3*-5) + 8

y = ??

Answer 39

-7

Answer 40

So your two solutions for the first question are

(4/3, 12) and (-5, 7)

Answer 41

thanks a lot

Answer 42

You're welcome!

Either my computer is lagging or it's QC so just tag me on your next post