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enot [183]
3 years ago
8

Pls guys help me with this i will give u more points!

Mathematics
1 answer:
gtnhenbr [62]3 years ago
5 0

3x - 4y =  - 6 \\  - 2x + y = 1

<em>The Ordered pair of solution will be (x,y)=(2,3)</em>

\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}

<h3>Steps below⤵️</h3>

  • Solve the second equation for y
  • Substitute the given value of y into the first equation
  • Solve the first equation for x
  • Substitute the given value of x into the second equation
  • Solve the equation for y
  • The possible solution of the system is the ordered pair (x,y)
  • And we are done solving!!~

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A jacket is advertised to be 85% off the original price. If the sale price is $102 what is the original price ?
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The answer would prob around $188.70 because .85x102=86.7 so add that to 102 and u get 188.70 but I'm prob wrong and I have a test on this tomorrow 


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Anybody Know This. ?
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Answer:

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Joe walked four over nine of a mile in four over five of an hour. What is his unit rate in miles per hour?
Irina-Kira [14]
<span>four over nine of a mile in four over five of an hour 

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</span>= 4                      5
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= 20
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Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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