Let us assume the first number to be = x
Let us assume the second number to be = y
Then
x^2 - y^2 = 5
x^2 = y^2 + 5
And
3x^2 - 2y^2 = 19
Multiplying the first equation by -2, we get
- 2x^2 + 2y^2 = - 10
Now subtracting the two equations, we get
x^2 = 9
x = 3
Putting the value of "x" in the first equation, we get
(3)^2 - y^2 = 5
9 - y^2 = 5
y^2 = 4
y = 2

I hope that the procedure is clear enough for you to understand and this is the answer that you were looking for.

5 0

3 years ago

Fill in the missing coefficients in the expansion of the binomial (2x^2+y^2)^4

Helga [31]

The expansion of the binomial:
( 2 x² + y² ) ^4 =
$(2 x^{2} ) ^{4} + 4 * (2 x^{2} ) ^{3}* y^{2} + 6 * ( 2 x^{2} ) ^{2}*( y^{2}) ^{2} +4*2 x^{2} *( y^{2}) ^{3}+( y^{2}) ^{4}$ =
$=16 x^{8}+32 x^{6} y^{2}+24 x^{4} y^{4} + 8 x^{2} y^{6}+ y^{8}$

6 0

3 years ago

Read 2 more answers

Simplify as much as possible: (7nm5o2) (-3m3o5)2

Yakvenalex [24]

Answer:

63m^11 no^12

Step-by-step explanation:

Dont know if thats o or and zero but I solve it by using o..if it was a zero then comment that so I can redo it anyways heres the explanation

(7nm^5 o^2) × (-3m^3 o^5)^2

7m^5 no^2) ×(-3m^3 o^5) ^2

(7m^5 no^2) × (3m^3 o^5)^2

7m^5 no^2 × (3m^3 o^5) ^2

7m^5 no^2 × 9m^6 o^10

63m^11 no^12

3 0

3 years ago