HSBC Equations Questions and Answers Quiz-1

From the second equation, put the value of 3^y in the first equation i.e.

3^y= (36/2)^2x

Therefore, we get

=> 2^x x (36/2)^2x =18

=> 2^(-x) = 18/36

=> 1/2^x = 1/2

=> 2^x = 2¹

=> x=1

Therefore, the value of x = 1.

 

Step-by-step explanation:

Consider the provided equation.

r = at^{2}

The equation can be written as:

a= \frac{r}{t^{2}}

Substitute the value of a in s = 2at

s= 2\times \frac{r}{t^{2}}\times t

s= 2\times \frac{r}{t}

s=\frac{2r}{t}

Hence, the required relation among s, r and a is s=\frac{2r}{t}

 

5log(a² bc^3)

5 loga² + 5logb + 5logc³

10log a + 5log b + 15log c

Let the no.be x
As in the question 'from'word is given we will take the given no.first
ATQ=5a²-3ab+7b²-x=a²+ab+b²
(its not in standard form so-)
5a²+7b²-3ab-x=a²+ab+b²
5a²-a²+7b²-b²-3ab-ab=x
4a²+6b²-4ab=x

Here,12+4 is 16 and it is written as 2⁴, then 12/4 is 3,
so now it is 2¹² so answer is 4096.

Discriminant = b² - 4ac

Given that a = 2k, b = 5k, c = 2

b² - 4ac = 0

(5k)² - 4(2k)(2) = 0

25k² - 16k = 0

k(25k - 16) = 0

k = 0 or 25k - 16 = 0

k = 0 or k = 16/25

Answer: k = 0 or k = 16/25

x + \frac{1}{x} = 4x^{3}+\frac{1}{x^{3}} = 52

Multiplying both side by x^{2} + \frac{1}{x^{2}}

\left (x^{3} + \frac{1}{x^{3}} \right ) \left(x^{2} + \frac{1}{x^{2}} \right) = 52 \left(x^{2} + \frac{1}{x^{2}} \right)

x^{5}+x+\frac{1}{x^{5}} +\frac{1}{x} = 52 \times 14x^{5}+\frac {1}{x^{5}} +4=728

x^{5} + \frac{1}{x^{5}}=728-4=724

For x= -2
First curve have the value y= 1 (-2,1)
Second curve have the value y = 7 ( -2,7)
For x= -1
First have (-1,5)
Second have ( -1,5)
For x = 0
First have (0,5)
Second have (0,5)
For x= 1
First have (1,7)
Second have (1,7)
So from the points it can be found that curve intersect at 3 points.
The two curves intersect thrice.

both have the same value
Packet-a : 1 gm cost = 48/100
Packet-b : 1 gm cost = 72/150