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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.
Answer:
25,000
Step-by-step explanation:
i think because 10 times 100 gets 1000 so now you have to do 250 times 100 to get 25,000
Answer:
49/8 is the value of k
Step-by-step explanation:
We have the system
x = -2y^2 - 3y + 5
x=k
We want to find k such that the system intersects once.
If we substitute the second into the first giving us k=-2y^2-3y+5 we should see we have a quadratic equation in terms of variable y.
This equation has one solution when it's discriminant is 0.
Let's first rewrite the equation in standard form.
Subtracting k on both sides gives
0=-2y^2-3y+5-k
The discriminant can be found by evaluating
b^2-4ac.
Upon comparing 0=-2y^2-3y+5-k to 0=ax^2+bx+c, we see that
a=-2, b=-3, and c=5-k.
So we want to solve the following equation for k:
(-3)^2-4(-2)(5-k)=0
9+8(5-k)=0
Distribute:
9+40-8k=0
49-8k=0
Add 8k on both sides:
49=8k
Divide both sides by 8"
49/8=k
In the equation for surface area the radius is squared ( r^2)
If the radius is doubled, the radius would be 2^2 more which equals 4
The surface area would be 4 times the old surface area.
