The parametric equations of a line are given below. $x = 1 + 3t$ $y = 4t$ Find the Cartesian equation of this line in the form $y = mx + c$. In how many different ways can all the letters of the word CONDOBOLIN be arranged in a line? Consider the polynomial $P(x) = x^3 + ax^2 + bx - 12$, where a and b are real numbers. It is given that $x + 1$ is a factor of $P(x)$ and that, when $P(x)$ is divided by $x - 2$, the remainder is -18. Find a and b.

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The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

Question 11

The parametric equations of a line are given below. $x = 1 + 3t$ $y = 4t$ Find the Cartesian equation of this line in the form $y = mx + c$. In how many diffe... show full transcript

Worked Solution & Example Answer:The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

Step 1

The parametric equations of a line are given below. $x = 1 + 3t$ $y = 4t$.

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Answer

To find the Cartesian equation in the form $y = mx + c$ , first express $t$ in terms of $x$ :

From the equation $x = 1 + 3t$ , rearranging gives: $t = \frac{x - 1}{3}$
Substitute $t$ into the equation for $y$ : $y = 4t = 4 \left(\frac{x - 1}{3}\right) = \frac{4(x - 1)}{3} = \frac{4}{3}x - \frac{4}{3}$

Thus, the Cartesian equation is: $y = \frac{4}{3}x - \frac{4}{3}$

Step 2

In how many different ways can all the letters of the word CONDOBOLIN be arranged in a line?

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Answer

The word CONDOBOLIN consists of 12 letters where:

C, N, D, B, L, I each occur once.
O occurs 2 times. Thus, the total number of arrangements is given by:

$\\text{Total Arrangements} = \frac{12!}{2!}$

Calculating:

Find $12!$ : 479001600
Find $2!$ : 2
So, $\\text{Total Arrangements} = \frac{479001600}{2} = 239500800$

Therefore, there are 239500800 different ways to arrange the letters.

Step 3

Consider the polynomial $P(x) = x^3 + ax^2 + bx - 12$, where a and b are real numbers.

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Answer

Since $(x + 1)$ is a factor, by the Factor Theorem: $P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 12 = 0$ $-1 + a - b - 12 = 0$ $a - b = 13 \\text{(Equation 1)}$
We also have: $P(2) = 2^3 + a(2)^2 + b(2) - 12 = -18 \to P(2) = -18$ $8 + 4a + 2b - 12 = -18$ $4a + 2b - 4 = -18$ $4a + 2b = -14 \to a + 0.5b = -3.5 \\text{(Equation 2)}$
Now solve for $a$ and $b$ using Equations 1 and 2: From Equation 1, $b = a - 13$ . Substitute into Equation 2: $a + 0.5(a - 13) = -3.5$ $a + 0.5a - 6.5 = -3.5$ $1.5a = 3 \to a = 2 \text{ and } b = -11$

The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

Worked Solution & Example Answer:The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

The parametric equations of a line are given below. $x = 1 + 3t$ $y = 4t$.

In how many different ways can all the letters of the word CONDOBOLIN be arranged in a line?

Consider the polynomial $P(x) = x^3 + ax^2 + bx - 12$, where a and b are real numbers.

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