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we have $70^2 + 61^2 + c^2 = d^2$ or $4900 +3721 + c^2 -= d^2$ or $d^2-c^2 = 8621$ or ($d+c)(d-c) = 8621$ to solve the same you can put 8621 as product of 2 numbers (one case 8621 * 1$ equiate one to d +c and another to d- c and solve for c and d. by choosing all pair of numbers fot erach pair...

kindly infrom what you have tried and where you are facing problems

Of what ?

yes. 4b – a + 1 = 0 as there is no x term. you need to divide by x+1 and get the constant term (putting x = -1 shall do also) this shall give 2 equations and you need to solve and proceed further;

In the above I find 100 solutions specified under the categories. $S(n)$

You are right. the question is easy. The product of numbers that are in P that are in Q cannot be more than the product of numbers that are in Q. this is so because the numbers in P that are in Q has to be be subset of numbers in Q. Now does Q contain at least one element that is not in P...

we have sum of 100 measurements = 2300 sum of 50 additional measurements = 27 * 50 = 1350 total sum = 2300 + 1350 = 2650 so arithmetic mean = $\frac{3650}{150} = \frac{73}{3}$ so B is bigger for a short cut 25 is mean of 23 and 27 more elements(100) have mean 23 and less (50) elements have...

There was a typo error in first line and I corrected the same. Otherwise I do not find error if any. This may be pointed

May be. I would like to know the correct answer

Width being less than square root of 600 I see that it is less than 25 so I have only one candidate d to check and width 10 so length 60 meets the criteria and hence the answer

For the profit we must have $R > C$ or $4t^2+t > 3t^2 + 9t$ Now you can proceed from here and find the result

Area of $\triangle STV$ = area of $\triangle UVT$ because they are on equal base and same base now area of $\triangle UVT$ is 3 times area of $\triangle UVX$ as height is same and base is 3 times so area of $\triangle STV$ = 3 * area of $\triangle UVX$ = $42cm^2$

In future kindly inform what you have tried and where you are stuck so that we can provide steps to proceed For this put $2^x = y$ and see what you get

you can write as E_{x_{y}} to get $E_{x_{y}}$

My Attempt After if left the 1st stop $\dfrac{5}{8}$ remained so $\dfrac{3}{8}$ was the number of more persons alighted than boarded The number is 180 so $\dfrac{3}{8}$ of number of persons when it left the $1^{st}$ stop is 180 so number of persons when it left $1^{st}$ stop...

because this is a challenge question you are required to answer it fully . this is not a question for help