Question

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Answer 1

Answer:

Q7:

Given the polynomial is :

$3 {x}^{2} - kx + 6$

and sum of zeros is = 3

we know that,

$\sf \: if \: a {x}^{2} + bx + c \: \: is \: a \: quadratic \\ \: polynomial \: \:and \: \alpha \: and \: \beta \: \: are \\ the \: zeroes \: \: \: then \: \\ \\ \sf \: \: sum \: of \: zeroes \: \: \: \alpha + \beta = - \frac{b}{a}$

In the question,

$\sf \: \alpha + \beta = 3 \\ \sf \: according \: to \: the \: question \: \\ \sf \: \: \: \: \: \: \: \: \: \alpha + \beta = - \frac{ - k}{3} = \frac{k}{3} \\ \sf \implies \: 3 = \frac{k}{3} \\ \sf \implies \: k = 3 \times 3 \\ \sf \therefore \: k = 9$

Q8:

Answer:

No.

Explanation :

We have $4^n$ , where n is a natural number.

so, n= 1,2,3,4...

$\sf \: if \: n = 1 \: \: then \: {4}^{1} = 4 \\ \sf \: if \: n = 2 \: \: then \: {4}^{2} = 16 \: \: \: and \: so \: on \:$

Again, if a number ends with zero then it is divisible by 5

Here, 4, 16 are not dividible by 5.

Therefore, $4^n$ can never end with zero.

Hence proved.