SOLUTION: A triangle has a base 12 centimeters and 8 centimeters. How much do you have to increase each dimension equally so that the area is tripled?
A man has a 50kg mixture that is one
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Question 550369: A triangle has a base 12 centimeters and 8 centimeters. How much do you have to increase each dimension equally so that the area is tripled?
A man has a 50kg mixture that is one-third cement and two-third sand. How much of a mixture that is half cement and half sand must be added to the 50kg mixture to produce a mixture that is one-fourth cement and three-fourth sand?
If a child were four times his present age, he would be half as old as his father now. If father is 30 yrs. old, how old is the child now?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
A triangle has a base 12 centimeters and 8 centimeters. How much do you have to increase each dimension equally so that the area is tripled?
I presume 8 is the height
area = 1/2 * 12 * 8
area = 48 cm^2
thrice = 48*3=144 cm^2
increase both by x
area after increase = 1/2 * (12+x)(8+x)
area = 1/2 * ( 96+12x+8x+x^2) by FOIL
area = 1/2 * (x^2+20x+96)
144= 1/2 ( (x^2+20x+96)
288=x^2+20x+196
x^2+20x-192=0
Find the roots of the equation by quadratic formula
a= 1 b= 20 c= -192
b^2-4ac= 400 + 768
b^2-4ac= 1168
= 34.18
x1=( -20 + 34.18 )/ 2
x1= 7.09
x2=( -20 -34.18 ) / 2
x2= -27.09
Ignore negative value
increase by 7.09 cm
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A man has a 50kg mixture that is one-third cement and two-third sand. How much of a mixture that is half cement and half sand must be added to the 50kg mixture to produce a mixture that is one-fourth cement and three-fourth sand?
mixture I 33.3 % cement 66.66% sand ------- 50 kgs
mixture II 50% cement , 50% sand to be mixed ---x
Total = 50+x
50*33.33+50*x= (50+x)*25
165.65+50x=1250+25x
25x=1250-165.65
25x=1084.35
x=43.3 kg of mix 2
If a child were four times his present age, he would be half as old as his father now. If father is 30 yrs. old, how old is the child now?
present age =x
if it was 4 times = 4x
4x=30
/4
x=7.5 years
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