Answer:
negative reciprocals of each other
U turn negative numbers from mixed numbers to improper fractions the same way u do with positive...u just temporarily disregard the negative sign.
Example :
- 2 2/3.....temporarily disregard the negative sign......take ur whole number (2) and multiply it by ur denominator (3) giving u 6, then add that to ur numerator (2) giving u 8....then put that number over the original denominator (3) giving u 8/3...and then put back ur negative sign...-8/3
There are 3 different numbers
slot method
how many 3 digit numbers?
321
312
123
132
213
231
6 numbes
how many 2 digit numbers?
12
21
13
31
23
32
6 numbers
how many 1 digit numbers
1
2
3
3 of them
6+6+3=15
15 different number, yah you ar right
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
