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:
yes
Step-by-step explanation:
D is the answer and here’s why
Check the picture below, so it reaches the maximum height at the vertex, let's check where that is
![h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}](https://tex.z-dn.net/?f=h%28t%29%3D64t-16t%5E2%2B0%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D%20%5C%5C%5C%5C%20h%28t%29%3D%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B-16%7Dt%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B64%7Dt%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B%2B0%7D%20%5Cqquad%20%5Cqquad%20%5Cleft%28-%5Ccfrac%7B%20b%7D%7B2%20a%7D~~~~%20%2C~~~~%20c-%5Ccfrac%7B%20b%5E2%7D%7B4%20a%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28-%5Ccfrac%7B%2064%7D%7B2%28-16%29%7D~~~~%20%2C~~~~%200-%5Ccfrac%7B%20%2864%29%5E2%7D%7B4%28-16%29%7D%5Cright%29%5Cimplies%20%5Cstackrel%7Bmaximum~height%7D%7B%282~~%2C~~%5Cstackrel%7B%5Cdownarrow%20%7D%7B64%7D%29%7D)
