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1/3 + 1/5 + 1/10
= (1/3)(30/30) + (1/5)(30/30) + (1/10)(30/30)
= 10/30 + 6/30 + 3/30
= (10 + 6 + 3)/30
= 19/30
Answer is 19/30

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7/8 of 100
= (7/8)*100
= 700/8
= 87.5 percent.

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160% = 160/100 = 1.6
Answer is 1.6

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Given,
y = 13 - 3x ----- (1)
2x + 9y = -8 ----- (2)
Substituting value of y in (2) from (1) we get
2x+9(13-3x) = -8
=> 2x+117-27x = -8
=> 117-25x = -8
=> 117-25x-117 = -8-117 ;(adding -117 to both sides)
=> -25x = -125
=> x = 5 ;(multiplying both sides by 1/-25)
Substituting the value of x in (1) we get
y = 13 - 3(5)
=>y = 13 - 15
=>y = -2
So, Answer is x=5, y=-2

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Anything raised to power 0 is 1.
so, 4^0 = 1

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principle amount p = 600
interest rate r = 1.4% = 0.014
time t = 30
So total interest after 30 months = p*r*t
= 600 * 0.014 * 30
= 252 dollars. Ans.

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X can install chairs in 10 hrs. So, in 1 hour X performs 1/10 part of the work.
Y can install chairs in 15 hrs. So, in 1 hour Y performs 1/15 part of the work.
X and Y together in 1 hour perform (1/10 + 1/15) = 1/6 part of the work.
1/6 part of the work is done in 1 hour
So, 1 (whole) work is done in 6 hours.

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Given,
w+x+y+z = 60
=> 1/2(x+y+z)+(x+y+z) = 60 ;[As w = 1/2(x+y+z)]
=> 3/2(x+y+z) = 60
=> x+y+z = 40 ;[multiplying both sides by 2/3]

Again
w+x+y+z = 60
=> w+40 = 60 ;[because x+y+z=40]
=> w = 20 ;[subtracting 40 from both sides]
Given
w+x+y+z = 60
=> (w+y+z)+1/3(w+y+z) = 60 ;[given x=1/3(w+y+z)]
=> 4/3(w+y+z) = 60
=> w+y+z = 45 ;[multiplying both sides by 3/4]
Again
w+x+y+z = 60
=> x+45 = 60 ;[because w+y+z = 45]
=> x = 15 ;[subtracting 45 from both sides]
Given
w+x+y+z = 60
=> (w+x+z)+1/4(w+x+z) = 60 ;[given y=1/4(w+x+z)]
=> 5/4(w+x+z) = 60
=> w+x+z = 48 ;[multiplying both sides by 4/5]
Again
w+x+y+z = 60
=> y+48 = 60 ;[because w+x+z = 48]
=> y = 12 ;[subtracting 48 from both sides]
We know,
w+x+y+z = 60
=> 20+15+12+z = 60 ;[as w=20, x=15, y=12]
=> z = 13
The solution is:
w=20, x=15, y=12 and z=13

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given
subtracted (-3)*x from both sides
multiplied both sides by 1/2
subtracted 1 from both sides
multiplied both sided by -1
Factoring left side yields
(x-3)(x-2) >= 0
The values of x for which x-2=0 or x-3=0 are x=2 and x=5. These points divide the coordinate line into three intervals,
(-inf,2], (2,3) and [3, +inf)
We need to check points of which of these three intervals give positive sign for the product (x-3(x-2). We shall choose arbitrary points on each of these intervals to determine the sign; these points are called test points. Lets say 1, 2.5 and 4 will be the test points for intervals (-inf,2], (2,3) and [3, +inf) respectively.
For interval (-inf,2] with test point 1 sign of (x-2)(x-3) is (-)(-) = +
For interval (2,3) with test point 2.5 sign of (x-2(x-3) is (+)(-) = -
For interval [3,+inf)with test point 4 sign of (x-2(x-3) is (+)(+) = +
The pattern of signs suggest that the solution set is
(-inf,2] U [3,+inf)