See, pls, the attached picture, the area determined by the conditions marked by yellow lines.
Answer: Adenike scored 64 marks, while Musa scored 45 marks
Step-by-step explanation: We shall start by assigning letters to each unknown variable. Let Adenike’s mark be d while Musa’s mark shall be m.
First of all, if Adenike obtained 19 marks more than Musa, then if Musa scored m, Adenike would score 19 + m (or d = 19 + m). Also if Adenike has obtained one and half her own mark (which would be 1 1/2d or 3d/2), it would have been equal to 6 times more than twice Musa’s mark (or 6 + 2m). This can be expressed as
3d/2 = 6 + 2m. So we now have a pair of simultaneous equations;
d = 19 + m ———(1)
3d/2 = 6 + 2m ———(2)
Substitute for the value of d into equation (2), if d = 19 + m
(3{19 + m})/2 = 6 + 2m
By cross multiplication we now have
3(19 + m) = 2(6 + 2m)
57 + 3m = 12 + 4m
We collect like terms and we have
57 - 12 = 4m - 3m
45 = m
We now substitute for the value of m into equation (1)
d = 19 + m
d = 19 + 45
d = 64
So Adenike scored 64 marks while Musa scored 45 marks
2 is the tens position meaning 2 tens
If the length of rectangular school hall is 4 times the width and the perimeter be 55m, then the length is 22m and the width of the school hall be 5.5m.
Given that the length of rectangular school hall is 4 times the width.
We are required to find the length and breadth of the rectangle which has the length be 4 times the width.
Perimeter of rectangle is the sum of all the length and breadth of that rectangles.
Perimeter of rectangle is given as 55m
Perimeter=2(L+B)
let the width of rectangle be x.
Length of that rectangle be 4x.
According to question,
2(x+4x)=55
2*5x=55
10x=55
x=55/10
x=5.5 m
Width be 5.5 m.
Length of rectangle be 4*5.5=22m.
Hence if the length of rectangular school hall is 4 times the width then the length is 22m and the width of the school hall be 5.5m.
Learn more about perimeter at brainly.com/question/19819849
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