# This are my questions: (1) A car covers 180m in (t-1) seconds and 324m in (t 3) seconds. ....

This are my questions:
(1) A car covers 180m in (t-1) seconds and 324m in (t 3) seconds. If it is traveling at a constant speed, calculate the value of t.
(2) Given that 1/m=t-1/n, make n the subject of the formula.
(3)  Find the dimensions of a rectangle whose perimeter and area are 46cm and 112cm square.
(4) A ship sails 5km due west and then 7km due south. Find correct to the nearest degree, its bearing from the original position.
(5) The semi -inter-quartile range of a distribution is 20. If the upper quartile is 96, find the lower quartile. (6)  solve the inequality: 3(2-3x) <1- 2x/3. (7) simplify :1/x 5- 2(x 2)/x square-25. straight answers is OK
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(1)
Since the car is traveling at constant speed, the speed after it travels
180 m is equal to the speed after it travels 324 m.

Speed = distance / time

speed 1 = 180m/(t-1)
speed 2 = 340m/(t + 3)
at constant speed, speed 1 = speed 2

180m/(t-1) = 340m/(t+3)
180(t+3) = 340(t-1) ....... cross multiplying
5(t + 3) = 9(t – 1) ......... dividing through by 36
5t + 15 = 9t - 9
15 + 9 = 9t - 5t
24 = 4t
t = 6

(2)
1/m = (t-1)/n
n = (t-1)m ...... cross multiplying eqn 1
n= (tm -m)

(3)
P = 2(l+w)
A = lw
where l = length of rectangle and w = width p = perimeter and a = area

46cm = 2l + 2w ..........eqn 1
divide eqn1 by 2
23cm = l + w
l = 23 - w ........eqn 3
112cm = lw ......... eqn 2
112cm = (23 - w)w
112cm = 23w - w^2
w^2 - 23w + 112 = 0

solving the quadratic gives w = 7 and w = 16
since the width of a rectangle is assumed to be shorter than the length,
if w = 7 , then  l = 23 - 7 = l = 16.
hence the dimensions are
l = 16 and w = 7

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