Problem 1 :
Find the area of the design as shown below.
Solution :
Area of rectangle = l x b
Area of semicircle = πr^{2}/2
Area of equilateral triangle = (√3/4)⋅a^{2}
= πr^{2}/2 + l x b + (√3/4)⋅a^{2}
= (1/2)⋅(22/7)⋅7^{2} + 20 ⋅ 14 + (√3/4)⋅14^{2}
= 441.368 cm^{2}
Problem 2 :
A cow is tied to one of the corners of a square-shaped field of sides of 10 m. The length of the rope is 7 m. How would one find the area of the field that the cow cannot graze?
Solution :
Area of square field = 10^{2}
Area of the filed that cow can graze = πr^{2}/4
= (1/4) ⋅ (22/7) ⋅ 7 ⋅ 7
= 77/2
Area of the filed that cow can graze = 100 - 77/2
= 61.5 m^{2}
Problem 3 :
Write the number 0.000001024 in scientific notation
Solution :
0.000001024 = 1.024 x 10^{6}
Problem 4 :
Write the number 1.423 × 10^{-6} in decimal form.
Solution :
= 1.423 × 10^{-6}
= 1.423/1000000
= 0.000001423
Problem 5 :
Perform the calculation and write the answer of the following in scientific notation.
(2000)^{2} ÷ (0.0001)^{4}
Solution :
(2000)^{2} ÷ (0.0001)^{4}
= (2x10^{3})^{2 }÷ (1x10^{-4})^{4}
= (4x10^{6}) ÷ (1x10^{-16}).
= 4x10^{6+16}
= 4x10^{22}
Problem 6 :
Solve for x
3 = log_{x}729
Solution :
x^{3} = 729
x^{3} = 3^{6}
x^{3} = (3^{2})^{3}
x = 9
So, the value of x is 9.
Problem 7 :
Change 2 = 64^{16} to logarithmic form.
Solution :
2 = 64^{16}
log_{64} 2 = 16
Problem 8 :
Evaluate
log_{9} (1/27)
Solution :
log_{9} (1/27)
= log_{9} (1/3)^{3}
= log_{9} 3^{-3}
= -1log_{9} 3
= -1/log_{3}9
= -1/log_{3}3^{2}
= -1/2log_{3}3
= -1/2
So, the answer is -1/2.
Problem 9 :
How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999 For example, both 121 and 211 have this property.
Solution :
110, 121, .........990
n = [(l-a)/d]+1
n = (990-110)/11 + 1
n = (880/11) + 1
n = 81
Number of three digit numbers divisible by 11 is 81.
Each number will consist of three digits, using these three digits we can make 6 possible 3 digit numbers.
81 x 6 = 486
In 486 numbers we have repeated counting.
For example, the number abc is divisible by 11, then cba is also divisible by 11. (Divisibility rule for 11)
So, 486/2 = 243
In those 243 numbers, we may have numbers consisting of one zero.
110, 220, 330, 440, 550, 660, 770, 880, 990
In this way, we can get 9 numbers. If we use the digits in the different places, that will not be divisible by 11 or it must a two digit number.
If the middle digit is 0, then
209, 307, 407, 506 by reversing the digits we will get 902, 703, 704, 605. So 8 numbers.
= 243 - (9+8)
= 226
Problem 10 :
Obtain the set builder representation of the set
A = {1, 1/2, 1/3, 1/4, 1/5, 1/6}
Solution :
A = {x: x = 1/n where n is a integer}
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