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ANTONII [103]
3 years ago
9

4x^-y^2 what and how would you solve this particular problem

Mathematics
1 answer:
dimulka [17.4K]3 years ago
5 0
It is x^4 - y^2  

so, expanding it, (x^2-y)(x^2+y)
 
hope it served
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2 years ago
X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
What is the domain for f(x) =26-2x?
anastassius [24]
Domain of f = R .............
7 0
3 years ago
PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP ME I BEGGING YOU PLZZZZZZZZZZZZZZZ HELP!!!! (55 POINTS)
Karo-lina-s [1.5K]

Answer:

1. 74 because 8(10) - 6

2.4(4) - 5(3) = 1

3. 7(3)+8(4) * 2 = 106

4. A

5.a

6. 20.5

7.26

8.a

9.a

10a

Step-by-step explanation:

3 0
3 years ago
-2(u+3)-5u<br><br> use distributive property to simplify
gayaneshka [121]
The answer is 
-2(u+3)-5u
 -2u-6-5u
  -3u-6

3 0
3 years ago
Read 2 more answers
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