Here, the cost of notebook = ₹x and the cost of pen = ₹y According to question, Cost of notebook = 2x Cost of Pen ⇒ x = 2y ⇒ RX-2y = 0 Here is the video explanation of this Question😊
Here, the cost of notebook = ₹x and the cost of pen = ₹y
According to question, Cost of notebook = 2x Cost of Pen
⇒ x = 2y ⇒ RX-2y = 0
(ii) Let p(x) = x⁴ + x³ + x² + x + 1 Putting x + 1 = 0, we get, x = -1 Using remainder theorem, when p(x) = x⁴ + x³ + x² + x + 1 is divided by x 1, reminder is given by p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 Since, reminder p(-1) ≠0, hence x + 1 is not a factor of x⁴ + x³ +Read more
(ii) Let p(x) = x⁴ + x³ + x² + x + 1
Putting x + 1 = 0, we get, x = -1
Using remainder theorem, when p(x) = x⁴ + x³ + x² + x + 1 is divided by x 1, reminder is given by p(-1)
= (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 – 1 + 1 – 1 + 1
= 1
Since, reminder p(-1) ≠0, hence x + 1 is not a factor of x⁴ + x³ + x² + x + 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen to be Rs y).
Here, the cost of notebook = ₹x and the cost of pen = ₹y According to question, Cost of notebook = 2x Cost of Pen ⇒ x = 2y ⇒ RX-2y = 0 Here is the video explanation of this Question😊
Here, the cost of notebook = ₹x and the cost of pen = ₹y
According to question, Cost of notebook = 2x Cost of Pen
⇒ x = 2y ⇒ RX-2y = 0
Here is the video explanation of this Question😊
See lessExpress the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: x-y/5-10 = 0
(ii) x- y/5 -10 = 0 ⇒ x- 1/5y - 10 = 0 Hence, here a = 1, b = -1/5 and c = -10.
(ii) x- y/5 -10 = 0
See less⇒ x- 1/5y – 10 = 0
Hence, here a = 1, b = -1/5 and c = -10.
Express the following linear equation in the form ax + by + c = 0 and indicate the values of a, b and c in case: –2x + 3y = 6
(iii) -2x + 3y = 6 ⇒ -2x + 3y - 6 = 0 Hence, here a = -2, b = 3 and c = -6.
(iii) -2x + 3y = 6
See less⇒ -2x + 3y – 6 = 0
Hence, here a = -2, b = 3 and c = -6.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. x¹⁰ + y³ + t⁵⁰
x¹⁰ + y³ + t⁵⁰, It is a polynomials in three variable as it contains three variable (x, y, t).
x¹⁰ + y³ + t⁵⁰, It is a polynomials in three variable as it contains three variable (x, y, t).
See lessDetermine which of the following polynomials has (x + 1) a factor: x⁴ + x³ + x² + x + 1
(ii) Let p(x) = x⁴ + x³ + x² + x + 1 Putting x + 1 = 0, we get, x = -1 Using remainder theorem, when p(x) = x⁴ + x³ + x² + x + 1 is divided by x 1, reminder is given by p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 Since, reminder p(-1) ≠0, hence x + 1 is not a factor of x⁴ + x³ +Read more
(ii) Let p(x) = x⁴ + x³ + x² + x + 1
Putting x + 1 = 0, we get, x = -1
Using remainder theorem, when p(x) = x⁴ + x³ + x² + x + 1 is divided by x 1, reminder is given by p(-1)
= (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 – 1 + 1 – 1 + 1
= 1
Since, reminder p(-1) ≠0, hence x + 1 is not a factor of x⁴ + x³ + x² + x + 1.
See here for Video explanation✌😁
See less