Solve the quadratic equation by factoring:<br><br> 6x²−x=15

. Given ????(5, −4) and T(−8,12):

damaskus [11]

Answer:

a) $y=\dfrac{13x}{16}-\dfrac{129}{16}$

b) $y = \dfrac{13x}{16}+ \dfrac{37}{2}$

Step-by-step explanation:

Given two points: $S(5,-4)$ and $T(-8,12)$

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: $y = mx + c$ where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

$m = \dfrac{12 - (-4)}{(-8) - 5}$

$m = \dfrac{-16}{13}$

to find the perpendicular of this gradient: we'll use:

$m_1m_2=-1$

both $m_1$ and $m_2$ denote slopes that are perpendicular to each other. So if $m_1 = \dfrac{12 - (-4)}{(-8) - 5}$ , then we can solve for $m_2$ for the slop of ther perpendicular!

$\left(\dfrac{-16}{13}\right)m_2=-1$

$m_2=\dfrac{13}{16}$ :: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope $m$ and the points $(x,y)$ . And plug into the equation: $(y - y_1) = m(x-x_1)$

side note: you can also use the $y = mx + c$ to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

$(y - y_1) = m(x-x_1)$

we have the slope of the perpendicular to ST i.e $m=\dfrac{13}{16}$

and the line should pass throught S as well, i.e $(5,-4)$ . Plugging all these values in the equation we'll get.

$(y - (-4)) = \dfrac{13}{16}(x-5)$

$y +4 = \dfrac{13x}{16}-\dfrac{65}{16}$

$y = \dfrac{13x}{16}-\dfrac{65}{16}-4$

$y=\dfrac{13x}{16}-\dfrac{129}{16}$

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for $T(-8,12)$

$(y - y_1) = m(x-x_1)$

$(y -12) = \dfrac{13}{16}(x+8)$

$y = \dfrac{13x}{16}+ \dfrac{104}{16}+12$

$y = \dfrac{13x}{16}+ \dfrac{37}{2}$

this is the equation of the line that is perpendicular to ST and passes through T

7 0

3 years ago

Two possible solutions of are -7 and 1. Which statement is true?

quester [9]

We can plug those values into the equation, and if the answer is incorrect, we'll know if either one is extraneous.

√11 - 2(-7) = √(-7)^2 + 4(-7) + 4

√25 = √25

5 = 5

The first solution, -7, makes the equation true, and so it is not extraneous.

√11 - 2(1) = √(1)^2 + 4(1) + 4

√9 = √9

3 = 3

The second solution, 1, makes the equation true, and so it is also not extraneous.

<h3>The correct option is D, neither solution is extraneous. </h3>

4 0

3 years ago

Solve the system of linear equations by substitution.<br><br> 4x−2y=14<br> y=0.5x−1

ANEK [815]

4x - 2(0.5x-1) =14
4x - 1x +2 = 14
3x + 2 = 14
X= 4
Plug that into second equation to get y

3 0

2 years ago

Simplify. Assume that no denominator is equal to zero.

nikklg [1K]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(2a^7b)^2}{32b^6}\implies \cfrac{(2^2a^{7\cdot 2}b^2)}{32b^6}\implies \cfrac{4a^{14}b^2}{32b^6}\implies \cfrac{a^{14}}{8b^6b^{-2}}\implies \cfrac{a^{14}}{8b^{6-2}}\implies \cfrac{a^{14}}{8b^4}$

8 0

3 years ago

Read 2 more answers

1. Three products from renewable resources are

Vesnalui [34]

Animals, wood, plants

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3 years ago

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Solve the quadratic equation by factoring: 6x²−x=15