Answer:
9 students neither speak hausa nor lgbo.
No circle can never be the perfect squares
Answer:
Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Step-by-step explanation:
Feets of fencing = 14
Perimeter = 14
Let x = Length y = width
Perimeter = 2(x + y)
For x = 1
2(1 + y) = 14
1 + y = 7
y =6
Area of rectangle ; x *y = 1 * 6 = 6 ft²
For x = 2
2(2 + y) = 14
2 + y = 7
y = 5
Area of rectangle ; x *y = 2 * 5 = 10 ft²
For x = 3
2(3 + y) = 14
3 + y = 7
y =4
Area of rectangle ; x *y = 3 * 4 = 12 ft²
For x = 4
2(1 + y) = 14
4 + y = 7
y = 3
Area of rectangle ; x *y = 4 * 3 =12 ft²
Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
200+30+0.2+0.05+0.001=230.251
<span>5t=3b+660....and ... 2t+5b=450
from the first we can see that t=(3b+660)/5 and using this value in the second...
(6b+1320)/5+5b=450 making all have common denominator..
6b+1320+25b=2250
31b=930
b=30...and since 2t+5b=450
t=150</span>
