Hey Aarya!
If we `scale` both the numerator and denominator by the same value, then we can create other equivalent fractions. By scale I mean: Multiply both the top and bottom of the fraction by the same value.
So here is one we can try.
Let's multiply numerator and denominator by 3.

Yay we did it! We found another fraction which is equivalent to 3/8.
Answer:

Step-by-step explanation:

Answer:
an average 7th grader owns 3 pairs of sneakers
we have

where
c---------> is the total cost of going to the carnival
t---------> is the number of
tickets purchased
we know that
<u>the domain of the function is the interva</u>l---------------> [0,∞)
Because the number of tickets can not be negative
<u>The range of the function is the interval</u>-------------> [10,∞)
Because the total cost can not be negative
using a graph tool
see the attached figure
The answer in the attached figure
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