Math help please :) (1 Viewer)

tonysoprano · Jul 5, 2015

Hey everyone. These are questions from the HSC papers. I've forgotten some really easy stuff so some help would be great please.

1. How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
a) 1
b) 2
c) 3
d) 4

2. Which expression is a factorisation of 8x^3 + 27?
(A) (2x – 3)(4x2 + 12x – 9)
(B) (2x + 3)(4x2 – 12x + 9)
(C) (2x – 3)(4x2 + 6x – 9)
(D) (2x + 3)(4x2 – 6x + 9)

3. Which equation represents the line perpendicular to 2x – 3y = 8, passing through the
point (2,0)?
(A) 3x + 2y = 4
(B) 3x + 2y = 6
(C) 3x – 2y = –4
(D) 3x – 2y = 6

I know they're easy. I'm just horrible at Math.

Drsoccerball · Jul 5, 2015

1) is 2 D: got this wrong in the hsc fml
2) D $(a+b)(a^2-ab+b^2) 3) Ceebs but find gradient of the line and times by -1 to find gradient and use point gradient formula$

tonysoprano · Jul 5, 2015

Drsoccerball said:
1) is 2 D: got this wrong in the hsc fml
2) D $(a+b)(a^2-ab+b^2) 3) Ceebs but find gradient of the line and times by -1 to find gradient and use point gradient formula[/QUOTE] The answer wasn't the issue, more so how to do it lmao.$

InteGrand · Jul 5, 2015

tonysoprano said:
Hey everyone. These are questions from the HSC papers. I've forgotten some really easy stuff so some help would be great please.

1. How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
a) 1
b) 2
c) 3
d) 4

2. Which expression is a factorisation of 8x^3 + 27?
(A) (2x – 3)(4x2 + 12x – 9)
(B) (2x + 3)(4x2 – 12x + 9)
(C) (2x – 3)(4x2 + 6x – 9)
(D) (2x + 3)(4x2 – 6x + 9)

3. Which equation represents the line perpendicular to 2x – 3y = 8, passing through the
point (2,0)?
(A) 3x + 2y = 4
(B) 3x + 2y = 6
(C) 3x – 2y = –4
(D) 3x – 2y = 6

I know they're easy. I'm just horrible at Math.

1. The answer is (B). This question was answered on this thread starting here: http://community.boredofstudies.org...unctions-multiple-choice-hsc.html#post6967946

$2. We use the factorisation rule $a^3+b^3 = (a+b)(a^2-ab+b^2)$, with $a=2x$ (so $a^3 = 8x^3$) and $b=3$ (so $b^3=27$), then we see that the answer is $(a+b)(a^2-ab+b^2)=(2x+3)((2x)^2-(2x)\cdot(3)+3^2)$, which simplifies to (D).$

$3. Rearranging the equation $2x-3y=8$, we have $3y=2x-8\Longleftrightarrow y = \frac{2}{3}x-\frac{8}{3}$, so the given line has slope $\frac{2}{3}$. Hence the required line has slope $-\frac{1}{\frac{2}{3}}=-\frac{3}{2}$. So the required line is of the form $y=-\frac{3}{2} x + b$. Substitute in the point (2,0), as we need this point to go through the line, to find $b$: $0=\frac{-3}{2}\times 2 + b\Rightarrow b = 3$. Hence the required line is $y=-\frac{3}{2}x + 3\Longleftrightarrow \frac{3}{2}x+y=3 \Longleftrightarrow 3x + 2y=6$, making (B) the answer.$

tonysoprano · Jul 5, 2015

InteGrand said:
1. The answer is (B). This question was answered on this thread starting here: http://community.boredofstudies.org...unctions-multiple-choice-hsc.html#post6967946

$2. We use the factorisation rule $a^3+b^3 = (a+b)(a^2-ab+b^2)$, with $a=2x$ (so $a^3 = 8x^3$) and $b=3$ (so $b^3=27$), then we see that the answer is $(a+b)(a^2-ab+b^2)=(2x+3)((2x)^2-(2x)\cdot(3)+3^2)$, which simplifies to (D).$

$3. Rearranging the equation $2x-3y=8$, we have $3y=2x-8\Longleftrightarrow y = \frac{2}{3}x-\frac{8}{3}$, so the given line has slope $\frac{2}{3}$. Hence the required line has slope $-\frac{1}{\frac{2}{3}}=-\frac{3}{2}$. So the required line is of the form $y=-\frac{3}{2} x + b$. Substitute in the point (2,0), as we need this point to go through the line, to find $b$: $0=\frac{-3}{2}\times 2 + b\Rightarrow b = 3$. Hence the required line is $y=-\frac{3}{2}x + 3\Longleftrightarrow \frac{3}{2}x+y=3 \Longleftrightarrow 3x + 2y=6$, making (B) the answer.$

Thanks a heap!

Drsoccerball · Jul 5, 2015

InteGrand said:
1. The answer is (B). This question was answered on this thread starting here: http://community.boredofstudies.org...unctions-multiple-choice-hsc.html#post6967946

$2. We use the factorisation rule $a^3+b^3 = (a+b)(a^2-ab+b^2)$, with $a=2x$ (so $a^3 = 8x^3$) and $b=3$ (so $b^3=27$), then we see that the answer is $(a+b)(a^2-ab+b^2)=(2x+3)((2x)^2-(2x)\cdot(3)+3^2)$, which simplifies to (D).$

$3. Rearranging the equation $2x-3y=8$, we have $3y=2x-8\Longleftrightarrow y = \frac{2}{3}x-\frac{8}{3}$, so the given line has slope $\frac{2}{3}$. Hence the required line has slope $-\frac{1}{\frac{2}{3}}=-\frac{3}{2}$. So the required line is of the form $y=-\frac{3}{2} x + b$. Substitute in the point (2,0), as we need this point to go through the line, to find $b$: $0=\frac{-3}{2}\times 2 + b\Rightarrow b = 3$. Hence the required line is $y=-\frac{3}{2}x + 3\Longleftrightarrow \frac{3}{2}x+y=3 \Longleftrightarrow 3x + 2y=6$, making (B) the answer.$

Hats down to integrand for always being bothered to provide pages of solutions for any question

snowflakeptical · Jul 5, 2015

Drsoccerball said:
Hats down to integrand for always being bothered to provide pages of solutions for any question

integrand's awesome! Joined in time for 2015ers

Math help please :) (1 Viewer)

tonysoprano

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Drsoccerball

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tonysoprano

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InteGrand

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tonysoprano

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Drsoccerball

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snowflakeptical

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