The area of the object with a shape of kite having diagonals of 14 inches and 10 inches is 70 in².
<h3>What is an equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
The area of kite = (product of diagonals)/2
Area of kite = (10 * 14) / 2 = 70 in²
The area of the object with a shape of kite having diagonals of 14 inches and 10 inches is 70 in².
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Subtract the following:
A) 18 rupees 9 paise from 75 rupees 80 paise.
B) 49 rupees 79 paise from 123 rupees 68 paise.
Answer
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Hint: Use decimal concept.
We know that, 1 rupee = 100 paise. We can reframe these questions as follows:
18 rupees 9 paise from 75 rupees 80 paise
18 rupees 9 paise can be represented as 18.09 rupees and 75 rupees 80 paise can be represented as 75.80 rupees. Now, on subtracting we’ll get,
75.80−18.09−−−−−− 57.71
Which means 57 rupees 71 paise
49 rupees 79 paise from 123 rupees 68 paise
49 rupees 79 paise can be represented as 49.79 rupees and 123 rupees 68 paise can be represented as 123.68 rupees. Now, on subtracting we’ll get,
123.68 −49.79−−−−−− 73.89
Which means 73 rupees 89 paise.
Note: We can also perform the subtraction by making all units the same that are paise and then subtract.
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D'ou know the area of a circle?
Answer:
A) A(t) = 4500*π - 1600*t
B) A(4) = 7730 in³
C) t = 8,8 sec
Step-by-step explanation:
The volume of the sphere is:
d max = 30 r max = 15 in
V(s) = (4/3)*π*r³ V(s) = (4/3)*π* (15)³
V(s) = 4500*π
A) Amount of air needed to fill the ball A(t)
A(t) = Total max. volume of the sphere - rate of flux of air * time
A(t) = 4500*π - 1600*t in³
B) After 4 minutes
A(4) = 4500*π - 6400
A(4) = 14130 - 6400
A(4) = 7730 in³
C) A(t) = 4500*π - 1600*t
when A(t) = 0 the ball got its maximum volume then:
4500*π - 1600*t = 0
t = 14130/1600
t = 8,8 sec