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saul85 [17]
3 years ago
7

By factoring and by completing the square

Mathematics
2 answers:
olga2289 [7]3 years ago
8 0
Ax^2 + bx + c

To factor, find two numbers that add up to your b term (-4) but also multiply to get your c term (-12)

That's how I got +2 and -6

2-6=-4

2 x -6 = -12

(x+2)(x-6)

Ahat [919]3 years ago
7 0
Factoring
what 2 numbers multiply to get -12 and add to get -4?
-6 and 2
(x-6)(x+2)=0
set to zero
x-6=0
x=6
x+2=0
x=-2

x=-2 and 6


complete the square
isolate x terms
(x^2-4x)-12=0
take 1/2 of -4 and square it ((-2)^2=4)
now add positive and neagtive inside
(x^2-4x+4-4)-12=0
factor perfect square
((x-2)^2-4)-12=0
get rid of parenthasees
(x-2)^2-4-12=0
(x-2)^2-16=0
add 16 both sides
(x-2)^2=16
sqrt both sides, don't forget positive an dneagit ve roots
x-2=+/-4
add 2 to both sides
x=2+/-4
x=2+4=6
x=2-4=-2

x=-2 and 6
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Pls answer it fast i will mark you as brainiest.
jeka94

\tt -1\dfrac{3}{4}=-\dfrac{(1\times 4)+3}{4}=-\dfrac{7}{4}

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2 years ago
The approximate distance between J and K is units. The approximate distance between K and L is units. The approximate distance b
Dima020 [189]
Count the distance d, which gives you 7.

Use Pythogarean theorem to find c:
{a}^{2}  +  {b}^{2}   =   {c}^{2}
Plug in the numbers:
{4}^{2}  +  {3}^{2}  =  {c}^{2}   \\  \\ = 16 + 9  \\  \\ = 25 \\  \\ c =  \sqrt{25}  = 5
c is 5.

Same thing with g:
{e}^{2}  +  {f}^{2}  =  {g}^{2}  \\  \\    {g}^{2} = {4}^{2}  +  {4}^{2}  \\  \\  = 16 + 16 \\  \\  = 32 \\  \\ g =  \sqrt{32} = 4 \sqrt{2}  = 5.66
g equals 5.66.

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3 years ago
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Vinil7 [7]
Isnt it 10 hours because he was charged 25$ and he is charged 1 dllr per hour soo
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8 0
3 years ago
A professional baseball team won 84 games this season. The team won 14 more games than it lost. There were no ties. How many gam
siniylev [52]

14+x=84

x=70

The team lost 70 games.

84+70=x

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The team played a total of 154 games.

5 0
3 years ago
Read 2 more answers
9in, 4in, 6in, _ is it a triangle?
gulaghasi [49]

Answer:

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Step-by-step explanation:

all sides are different

Brainlist Pls!

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