1
Simplify \frac{1}{2}\imath n(x+3)21ın(x+3) to \frac{\imath n(x+3)}{2}2ın(x+3)
\frac{\imath n(x+3)}{2}-\imath nx=02ın(x+3)−ınx=0
2
Add \imath nxınx to both sides
\frac{\imath n(x+3)}{2}=\imath nx2ın(x+3)=ınx
3
Multiply both sides by 22
\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2
4
Regroup terms
\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı
5
Cancel \imathı on both sides
n(x+3)=nx\times 2n(x+3)=nx×2
6
Divide both sides by nn
x+3=\frac{nx\times 2}{n}x+3=nnx×2
7
Subtract 33 from both sides
x=\frac{nx\times 2}{n}-3x=nnx×2−3
Answer:
16 m × 11 m
Step-by-step explanation:
The dimension of the gym is 20 m x 15 m. An outbound of 2m width is to be cut out from the gym to form the basketball court.
The original length of gym = 20 m and original width of gym = 15 m
2 m would be cut at both sides of the gym length for the outbound. Also 2 m would be cut at both sides of the gym width for the outbound. Therefore:
Length of basketball court = 20 m - (2 * 2m) = 20 m - 4 m = 16 m
Width of basketball court = 15 m - (2 * 2m) = 15 m - 4 m = 11 m
Therefore the dimensions of the basketball court is:
16 m × 11 m
A cycle, that is in the consideration of the earth and also help the combinations of things
