Selling price = $110
Discount % = 45%
Let the original/marked price be x.
⇒x - (45% of x) = 110
⇒x - 45x/100 = 110
⇒100x - 45x = 11000
⇒55x = 11000
⇒x = 200
Thus, the original price was $200.
Area of rhombus = 1/2 * product of diagonals
= 1/2 * 15 * 8
= 60 sq. ft.
Total sum of angles in heptagon = 1080
⇒132 + 123 + 131 + 125 + x + 124 + 129 = 1080
⇒764 + x = 1080
⇒x = 316º
In the given cuboid,
l = 8; b = 4; h = 15
Surface area of cuboid = 2(lb + bh + hl)
= 2(8*4 + 4*15 + 15*8)
= 2(32 + 60 + 120)
= 2 * 212
= 424 sq. yds
3. Consider a right angled triangle ABC,
Let AC (length of string from ground) = 25 m ...[AC is hypotenuse]
BC (person to barn distance) = 18 m
cos∠ACB = 18/25
= 0.72
∠ACB = 43.95º
Thus, the string makes 43.95º with the ground.
In △ABC,
∠ABC = 90°
∵ BM bisects ∠ABC
∴ ∠ABM = 1/2∠ABC
= 1/2 * 90
= 45°
cos∠ABM = cos 45
= \({1 \over{ \sqrt{2}}}\)
No, this answer was wrong. I've posted the correct one.
Turns out, the previous answer I posted was wrong.
Here's the correct solution :-
z2 = 24 - 32i
Let z = x + iy
⇒ z2 = x2 + i2y2 + 2xiy
= x2 - y2 + 2xiy
Now, the real part,
x2 - y2 = 24 ...(1)
Imaginary part,
2xy = -32
xy = -16
x = -16/y
From (1)
x2 - 256/x2 = 24
x4 - 256 = 24x2
x4 - 24x2 - 256 = 0
Solving this equation,
x2 = -8, 36
⇒x = 6
y = -8/3
Thus z = 6 - 8i/3
|z| = \({ \sqrt{6^2 + (8/3)^2}}\)
= 6.57
Hope this helps :)
PQRS is a cyclic quadrilateral.
Since opp. angles are supplementary,
⇒∠P + ∠R = 180
⇒5y + 15 + 15y + 45 = 180
⇒20y + 60 = 180
⇒20y = 120
⇒y = 6
Thus,
∠P = 45°
∠Q = 111°
∠R = 135°
∠S = 360 - (45 +111+135) [Sum of all angles = 360°]
= 69º
z = 24 -32i
|z| = \({\sqrt{24^2 + 32^2}}\)
= \({\sqrt{1600}}\)
= 40
