To solve this you want to plug in your x's and y's to see if they match.
A. 3(11)-4=33-4=29 and 29 is not equal to 5 so A is not a solution
B.3(5)-4=15-4=11=11 3(3)-4=9-4=5 and 5 is not equal to 2 so B is not a solution
C. 3(2)-4= 6-4=2 2 is not equal to 3 so C is not a solution
D is the answer. 3(5)-4=11 and 3(2)-4= 2
There is no square root of five...
Answer:
{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x
4
+4x
3
y+6x
2
y
2
+4xy
3
+y
4
Step-by-step explanation:
1 Use Square of Sum: {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}(a+b)
2
=a
2
+2ab+b
2
.
({x}^{2}+2xy+{y}^{2})({x}^{2}+2xy+{y}^{2})(x
2
+2xy+y
2
)(x
2
+2xy+y
2
)
2 Expand by distributing sum groups.
{x}^{2}({x}^{2}+2xy+{y}^{2})+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x
2
(x
2
+2xy+y
2
)+2xy(x
2
+2xy+y
2
)+y
2
(x
2
+2xy+y
2
)
3 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x
4
+2x
3
y+x
2
y
2
+2xy(x
2
+2xy+y
2
)+y
2
(x
2
+2xy+y
2
)
4 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}({x}^{2}+2xy+{y}^{2})x
4
+2x
3
y+x
2
y
2
+2x
3
y+4x
2
y
2
+2xy
3
+y
2
(x
2
+2xy+y
2
)
5 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}{x}^{2}+2{y}^{3}x+{y}^{4}x
4
+2x
3
y+x
2
y
2
+2x
3
y+4x
2
y
2
+2xy
3
+y
2
x
2
+2y
3
x+y
4
6 Collect like terms.
{x}^{4}+(2{x}^{3}y+2{x}^{3}y)+({x}^{2}{y}^{2}+4{x}^{2}{y}^{2}+{x}^{2}{y}^{2})+(2x{y}^{3}+2x{y}^{3})+{y}^{4}x
4
+(2x
3
y+2x
3
y)+(x
2
y
2
+4x
2
y
2
+x
2
y
2
)+(2xy
3
+2xy
3
)+y
4
7 Simplify.
{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x
4
+4x
3
y+6x
2
y
2
+4xy
3
+y
4
Answer:
a) 1/2
b) 6/100
c)1/4
I'll explain how I got these answers later.
Its wrong for x = 2
y = (3/4) * 2^2 = 3 (NOT 4)
