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Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
$S _{22}$, if $d =22$ and $a _{22}=149$
→$S _{22}$, if $d =22$ and $a _{22}=149$
For what value of $k$, is -3 a zero of the polynomial $x^2+11 x+k$ ?
→Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
Solve the following simultaneous equations.
$\frac{7}{2 x +1}+\frac{13}{ y +2}=27 ; \frac{13}{2 x +1}+\frac{7}{ y +2}=33$
→$\frac{7}{2 x +1}+\frac{13}{ y +2}=27 ; \frac{13}{2 x +1}+\frac{7}{ y +2}=33$
Solve the following quadratic equations by factorization:
$5x^2 - 3x - 2 = 0$
→$5x^2 - 3x - 2 = 0$
The areas of two similar triangles are $169\ m^2$ and $121\ cm^2$ respectively. If the longest side of the larger triangle is $26cm$, what is the length of the longest side of the smaller triangle?
→If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - x - 2$, find the value of $\frac{1}{\alpha}-\frac{1}{\beta}$
→A box contains $5$ strawberry chocolates, $6$ coffee chocolates and $2$ peppermint
chocolates. If one of the chocolates is picked from the box at randomly, Find the
probability of the following events by completing the activity.
(i) Event A : it is a coffee chocolate.
(ii) Event B : it is a peppermint chocolate.Activity : Let 'S' is the sample space.
$\therefore n(S) = 5+6+2 = 13$
(i) Event A : it is a coffee chocolate
$\therefore n(A) =$ …………
$\therefore P(A) = \frac{\square}{n(S)}$ ………… Formula
$\therefore P(A)=\frac{\square}{13}$
(ii) Event B : it is a peppermint chocolate.
$\therefore n(B) =$ …………
$\therefore P(B) =\frac{\square}{n(S)}$ ………… Formula
$\therefore P(B) =\frac{\square}{13}$
→chocolates. If one of the chocolates is picked from the box at randomly, Find the
probability of the following events by completing the activity.
(i) Event A : it is a coffee chocolate.
(ii) Event B : it is a peppermint chocolate.Activity : Let 'S' is the sample space.
$\therefore n(S) = 5+6+2 = 13$
(i) Event A : it is a coffee chocolate
$\therefore n(A) =$ …………
$\therefore P(A) = \frac{\square}{n(S)}$ ………… Formula
$\therefore P(A)=\frac{\square}{13}$
(ii) Event B : it is a peppermint chocolate.
$\therefore n(B) =$ …………
$\therefore P(B) =\frac{\square}{n(S)}$ ………… Formula
$\therefore P(B) =\frac{\square}{13}$
The adjoining pie-diagram shows the percentage distribution of the expenditure incurred in publishing a book. Study the diagram and answer the questions:
(a) If for certain quantity of books, the publisher has to pay ₹ 30,600 as printing, then what will be the amount of royalty to be paid for these books?
(b) What is the central angle of the sector corresponding to the expenditure incurred on Royalty?
(a) If for certain quantity of books, the publisher has to pay ₹ 30,600 as printing, then what will be the amount of royalty to be paid for these books?
(b) What is the central angle of the sector corresponding to the expenditure incurred on Royalty?
If the HCF of 657 and 963 is expressible in the form 657x + 963x - 15, find x.