All Answers

+15127

+1

kg/(lenght x thickness x width) = 62kgm3 can i know the weight at least
how to count

$\delta =62\ kg/m^3 $

$\delta \ is \ the \ density.\\ If\ you \ know \ length, \ thickness \ width, the \ mass \ is $

$m= \delta\times ( length\times \ thickness\times \ width)\\=62\ kg/m^3\times( length\times \ thickness\times \ width)m^3=m \ kg$

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asinusApr 25, 2017

#1

+33661

+2

Using the left-hand endpoint, the approximate area is f(a)*(b - a) = 5*4 → 20

Using the right-hand endpoint, the approximate area is f(b)*(b - a) = 13*4 → 52

Using the midpoint, the area is (f(a) + f(b))/2 * (b - a) = (5 + 13)/2 * 4 → 9*4 → 36

(Not clear why you mentioned 4 rectangles!).

AlanApr 25, 2017

#2

+33661

+3

If we assume this is meant to be cos θ = 0.8, then sin θ = sqrt(1-0.8^2) → 0.6 and tan θ = 0.6/0.8 → 0.75

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AlanApr 25, 2017

#1

+33661

+2

Let N be the original amount.

clothes = N/4

shoes = (1/3)*(3N/4) → N/4

bag = N/4 + 15

Total spent = 3N/4 + 15 so balance = N - (3N/4 + 15) → N/4 + 15

but this equals 95, so:

N/4 + 15 = 95

N/4 = 80

N = 320

Ahmad started with RM 320

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AlanApr 25, 2017

#1

+33661

+3

Calculate the ratio of the (n+1)'th term to the n'th term and then let n tend to infinity. You will find in each case that the ratio tends to 1, so that there are an infinite number of same size terms being added, leading to an infinite sum.

For example, the ratio of the (n+1)'th to the n'th term for your first sum is: (n+2)(n^2+2n+3)/[(n+1)(n^2+4n+6)] which tends to 1 as n tends to infinity.

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AlanApr 25, 2017

#1