\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{71}{97}\)
\(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{3}{5}\) , \(\frac{71}{97}\)
\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
\(\frac{9}{16}\), \(\frac{3}{5}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
Correct answer is B
\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73
Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\)
Multiply (x + 3y + 5) by (2x2 + 5y + 2)
2x2 + 3yx2 + 10xy + 15y2 + 13y + 10x2 + 2x + 10
2x3 + 6yx2 + 5xy + 15y2 + 31y + 5x2 + 2x + 10
2x3 + 6xy2 + 5xy + 15y2 + 12y + 10x2 + 2x = 10
2x2 + 6xy2 + 5xy + 15y2 + 13y + 10x2 + 2x + 10
2x3 + 2yx2 + 10xy + 10y2 + 31y + 5x2 + 10
Correct answer is B
\((x + 3y + 5)(2x^{2} + 5y + 2)\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)
3\(\frac{2}{3}\)
5\(\frac{1}{4}\)
6\(\frac{1}{2}\)
8
8\(\frac{1}{8}\)
Correct answer is A
\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)
= \(4\frac{29}{24}\)
\(\equiv 5\frac{5}{24}\)
\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)
= \(\frac{40 - 3}{24}\)
= \(\frac{37}{24}\)
\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)
= \(\frac{88}{24}\)
= 3\(\frac{2}{3}\)
15sq.cm
20sq.cm
13sq.cm
16sq.cm
17sq.cm
Correct answer is A
Area of the rectangular picture = L x B = 8 x 6
= 48 sq.cm.
Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm
area of the frame is 63 - 48
= 15 sq. cm
30o
60o
45o
25o
Correct answer is B
Let x rep. the angle of depression of the foot of the tree.
tan 30o = \(\frac{y}{100}\)
y = 100 tan 30o
= 57.8
By Pythagoras, AC2 = 3002 + 582
= 900 + 3354
tan x = \(\frac{opp}{adj}\)
= \(\frac{58}{300}\)
= 0.19
tan x = 0.19
x = tan 0.19
= 60o