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SA1 Answers2

SA1 Answers2

SECTION - B

9. Here, n is any natural number so let us plug

n = 1, we get,

\(4^{n}\) = 4

n = 2, we get,

\(4^{2}\) = 16

Ending with 6.

n = 3, We get,

\(4^{3}\) = 64

Ending with 4

Therefore \(4^{n}\) where 'n' is a natural number can end with either 4 or 6 but not 0.



10)

x² + 7x + 10 = x² + 2x + 5x + 10

                    = x(x+2) + 5(x+2)

                    = (x+2)(x+5)

Therefore the zeroes are -2 and -5

Sum of the roots = -2 - 5 = -7

Sum of the roots = \(\large \frac{-7}{1}\) = -7

Product of the roots = -2(-5)

                              = 10

Product of the roots = \(\large \frac{10}{1}\)

                              = 10


11)

\( \large \frac{3}{2k-1}\) \(\neq\) \(\large \frac{1}{k-1}\) 

3(k-1) \(\neq\) 2k-1

3k - 3 \(\neq\) 2k - 1

k \(\neq\) 2

Therefore the value of k is 2.


12)

\(\angle B =90°\)

Using Pythagoras theorem,

AC² = AB² + BC²         ---------(1)

AD² = AB² + BC² + CD²

       = AC² + CD²         (\(\because from (1)\))

Using converse of Pythagoras theorem we get \(\angle ACD = 90°\).

Hence proved.


PQRT forms a rectangle,So

PT = QR = 5cm
PR = 13cm
PS = 14cm
TS = PS - PT = 14 - 5 = 9cm

Using Pythagoras theorem we get 

TR = √(PR² - PT²)

     = √(169 - 25)

     = √144

     = 12

Tan S = \(\large \frac{TR}{TS}\)

         = \(\large \frac{12}{9}\)

         = \(\large \frac{4}{3}\)

SA1 Answers2


13)

sinΘ = \(\large \frac{3}{5}\) = \(\large \frac{AB}{AC}\)

AB = 3
AC = 5
BC = √(AC² - AB²)
     = √(25 - 9)
     = √16
     = 4

cosΘ = \(\large \frac{4}{5}\)


14)

median = l + \(\large \left [ \frac{\frac{n}{2}-C.F}{f}\right]\)h

Here, l = 125 , f = 20 , C.F. = 22, median = 137, h = 145-125 = 20

175 = 125 + \(\large \left[ \frac{\frac{n}{2}-22}{20}\right]\)*20

175 = 125 + \(\large \frac{n}{2}\) - 22

175 = 103 + \(\large \frac{n}{2}\)

\(\large \frac{n}{2}\) = 175 - 103

      = 72

n = 72*2

   = 144

Therefore the sum of frequencies (n) is 144.


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