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Soloha48 [4]
2 years ago
13

Select the correct answer from each drop-down menu. A system of equations and its solution are given system A x-y=7 -3x+9y=-39 s

olution: (4,-3) complete the sentences to explain what steps were followed to obtain the system of equations below. system B x-y=7 6y=-18
Mathematics
1 answer:
galina1969 [7]2 years ago
4 0

Answer:

Step-by-step explanation:

x-y=7

-3x+9y=-39

Divide the second equation by 3

-x +3y = -13

Add this to the first equation

x-y=7

-x +3y = -13

----------------------

0x +2y = -6

Divide by 2

2y/2 = -6/2

y = -3

Now find x

x-y = 7

x -(-3) = 7

x+3 = 7

Subtract 3 from each side

x = 4

(4,-3)

Or by substitution

x-y=7  

solve for x

x = 7+y

-3x+9y=-39

Substitute y+7 in for x

-3(7+y) +9y = -39

Distribute

-21 -3y +9y = -39

Combine like terms

-21 +6y = -39

Add 21 to each side

6y = -18

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-3x^{2}-21x-54 what are the zeros? (Solutions)
Irina-Kira [14]
The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54

remember if we can do
xy=0 then assume x and y=0

so factor

0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is

if you have
ax^2+bx+c=0 then
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

x^2+7x+18
a=1
b=7
c=18

x=\frac{-7+/- \sqrt{7^{2}-4(1)(18)} }{2(1)}
x=\frac{-7+/- \sqrt{49-72} }{2}
x=\frac{-7+/- \sqrt{-23} }{2}
i=√-1
x=\frac{-7+/- i\sqrt{23} }{2}



the zerose would be
x=\frac{-7+ i\sqrt{23} }{2} or \frac{-7- i\sqrt{23} }{2}




4 0
2 years ago
I GIVE BRAINLIEST PLS HELP
ioda
The answer is 8
Here's why:
{ ( \frac{( {6}^{7}) \times ( {3}^{3})  }{(  {6}^{6}) \times ( {3}^{4}  ) } )}^{3}  =   \\ ( \frac{6}{3} ) ^{3}  =  \\ \frac{216}{27}  = 8
The exponents are subtracted one from another when divided.
\frac{ a ^{b} }{ {a}^{c} } =  {a}^{b - c}
We can look at the problem this way:
( \frac{6^{7} }{6 ^{6} }  \times  \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6}  \times  {3}^{3 - 4} ) =  \\ ({6}^{1}  \times  {3}^{ - 1} )
Since we have the power of -1 on the 3, we apply this rule:
{a}^{ - b}  =  \frac{1}{ {a}^{b} }
Also this rule because we have the power of 1 on the 6:
{a}^{1}  = a
Then we get this:
(6 \times  \frac{1}{3} )^{3}  = ( \frac{6}{3} )^{3}
We apply the rule:
( \frac{a}{b}) ^{c}   =  \frac{ {a}^{c} }{ {b}^{c} }
We get this:
\frac{{6}^{3} }{ {3}^{3} } =  \frac{216}{27}  = 8
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