Maths-
What is 8 in roman numeral
Maths-General
- Vii
- iiv
- Viii
- Vi
Answer:The correct answer is: ViiiV= 5 adding 3 times I will be 8
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Related Questions to study
Maths-
Find the area of a given triangle whose perimeter is equal to 26cm .10 cm
Perimeter= 26cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Find the area of a given triangle whose perimeter is equal to 26cm .10 cm
Maths-General
Perimeter= 26cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Height=10cm
Hypotenuse=12cm
Area=?
Perimeter of triangle=sum of sides
26=10+12+ b
26=22+b
26-22=b
4cm=b
Area=1/2*base*height
=1\2 * 4*10= 20 sq. cm
Maths-
Himani saves Rs 700 a month and pratibha saves Rs 900 a month. Find the ratio in their savings.
Himani’s ratio = Rs 700 per month
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Himani saves Rs 700 a month and pratibha saves Rs 900 a month. Find the ratio in their savings.
Maths-General
Himani’s ratio = Rs 700 per month
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Pratibha’s ratio = Rs 900 per month
Ratio of their savings = 700/900
= (700/100) / (900/100)
=7/9
= 7:9
Maths-
11 divided by 1/3
11 ÷1/3 = 11 x 3/1 = 33
11 divided by 1/3
Maths-General
11 ÷1/3 = 11 x 3/1 = 33
Maths-
What is 8 x 1/32
8x 1/32 = 8/32 = ¼
What is 8 x 1/32
Maths-General
8x 1/32 = 8/32 = ¼
Maths-
The area of a rectangular park is 1500 sq m. If the lenth is 50m, find its width. Also find the perimeter of the park.
Area of rectangular park = 1500 sq m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
The area of a rectangular park is 1500 sq m. If the lenth is 50m, find its width. Also find the perimeter of the park.
Maths-General
Area of rectangular park = 1500 sq m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Length = 50m
Bredth of the park= (1500/50)
= 30m
Perimeter =2*(l+b)
= 2*(50 + 30)
=2* 80
= 160m
Maths-
Find the value of x:
x+60=180 {linear angle sum property}
x= 180-60
x = 120
x= 180-60
x = 120
Find the value of x:
Maths-General
x+60=180 {linear angle sum property}
x= 180-60
x = 120
x= 180-60
x = 120
Maths-
iv in roman number is
V= 5 , I = 1
We subtract the smaller value in the left, therefore 5-1 = 4
We subtract the smaller value in the left, therefore 5-1 = 4
iv in roman number is
Maths-General
V= 5 , I = 1
We subtract the smaller value in the left, therefore 5-1 = 4
We subtract the smaller value in the left, therefore 5-1 = 4
Maths-
What is LCM of 4 and 6.
LCM of 4 & 6= 2*2*3 = 12
What is LCM of 4 and 6.
Maths-General
LCM of 4 & 6= 2*2*3 = 12
Maths-
There are two circle with same centre O. The radius of larger circle is 4cm and that of the smaller circle is 2cm.
Find area of the shaded region.
Find area of the shaded region.
Ar of shaded region =ar. of larger circle – ar. of smaller circle
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
There are two circle with same centre O. The radius of larger circle is 4cm and that of the smaller circle is 2cm.
Find area of the shaded region.
Find area of the shaded region.
Maths-General
Ar of shaded region =ar. of larger circle – ar. of smaller circle
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
{ar. of circle = 22/7*r*r}
= 22/7*4*4-22/7*2*2
=22/7(16-4)
=22/7*12
=37.7 sq cm
Maths-
Least common multiple for 12 and 5 is
As 12 and 5 are co-prime, there product will be the LCM
Least common multiple for 12 and 5 is
Maths-General
As 12 and 5 are co-prime, there product will be the LCM
Maths-
If all the sides of triangle ,so the triangle is called __________ triangle:
If all the sides of a triangle is equal, then its called an equilateral
Triangle.
Triangle.
If all the sides of triangle ,so the triangle is called __________ triangle:
Maths-General
If all the sides of a triangle is equal, then its called an equilateral
Triangle.
Triangle.
Maths-
What is 9 divided by 72
9/72, dividing Numerator and denominator by 9 we get 1/8
What is 9 divided by 72
Maths-General
9/72, dividing Numerator and denominator by 9 we get 1/8
Maths-
12 divided by 26
12/26, dividing Numerator and denominator by 2 we get 6/13
12 divided by 26
Maths-General
12/26, dividing Numerator and denominator by 2 we get 6/13
Maths-
-3/-8 is
Positive rational number
-3/-8 is
Maths-General
Positive rational number
Maths-
¼ litre of milk cost Rs 15/2 . Find the cost of 9/2 litre of milk.
Cost of ¼ litre milk= Rs 15/2
Cost of 1 litre milk =Rs (15/2) /(1/4)
=Rs (15/2 *4/1)
= Rs 30
Cost of 9/2 litre of milk =30 * 9/2
= (30*9)/2
= Rs 135
Cost of 1 litre milk =Rs (15/2) /(1/4)
=Rs (15/2 *4/1)
= Rs 30
Cost of 9/2 litre of milk =30 * 9/2
= (30*9)/2
= Rs 135
¼ litre of milk cost Rs 15/2 . Find the cost of 9/2 litre of milk.
Maths-General
Cost of ¼ litre milk= Rs 15/2
Cost of 1 litre milk =Rs (15/2) /(1/4)
=Rs (15/2 *4/1)
= Rs 30
Cost of 9/2 litre of milk =30 * 9/2
= (30*9)/2
= Rs 135
Cost of 1 litre milk =Rs (15/2) /(1/4)
=Rs (15/2 *4/1)
= Rs 30
Cost of 9/2 litre of milk =30 * 9/2
= (30*9)/2
= Rs 135