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3 years ago

Line E passes through the points (2, 3) and (4, -3). What is the slope of a line perpendicular to line E?

1 answer:

7 0

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 2 &,& 3~) 
% (c,d)
&&(~ 4 &,& -3~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-3}{4-2}\implies \cfrac{-6}{2}\implies \cfrac{-3}{1}$

now, a line perpendicular to that one, will have a "negative reciprocal" slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for}\quad \cfrac{-3}{1}\\\\
negative\implies +\cfrac{3}{ 1}\qquad reciprocal\implies +\cfrac{ 1}{3}\implies \cfrac{1}{3}$

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4 years ago

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3 years ago

How on earth do u do this?

11111nata11111 [884]

Answer:

ΔABC ≅ ΔZXN

Step-by-step explanation:

Find the letters of the matching parts of the triangles.

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C is the remaining angle; as is N.

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ΔABC ≅ ΔZXN

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4 years ago

H(x)=-x^2+6x. So what is the value of h(2)

crimeas [40]

✩ Answer:

✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*

$\bold{Hello!}\\\bold{Your~answer~is~below!}$

✩ Step-by-step explanation:

✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*

✺ Quadratic polynomials can be factored using the transformation $ax^2+bx+c=a(x-x_{1})(x-x_{2} )$ , where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^2+bx+c=0$ :

$-x^2+6x=0$

✺ All equations of the form $ax^2+bx+c=0$ can be solved using the quadratic formula:

$-b=\frac{+}\\\sqrt{b^2-4ac}\\~~~~~~~~~~~~~2a$

✺ The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction:

$x=\frac{\sqrt{-6\frac{+}\\\sqrt{6^2}}}{2(-1)}$

✺ Take the square root of $6^2$ :

$x=\frac{\sqrt{-6\frac{+}\\{6}}}{2(-1)}$

✺ Multiply $2$ times $-1$ :

$x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$

✺ Now solve the equation $x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$ when ± is plus. Add $-6$ to $6$ :

$x=\frac{0}{-2}$

✺ Divide $0$ by $-2$ :

$x=0$

-OR-

✺ Now solve the equation $x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$ when ± is minus. Subtract $6$ from $-6$ :

$x=\frac{-12}{-2}$

✺ Divide $-12$ by $-2$ :

$x=6$

✺ Optional : Factor the original expression using $ax^2+bx+c=a(x-x_{1})(x-x_{2} )$ . Substitute $0$ for $x_{1}$ and $6$ for $x_{2}$ :

$-x^2+6x=-x(x-6)$

✩ Answer:

✺ Factored Form: $x(x-6)$

✺ Exact Form: $x=6$

✺ Graph Point Form: $x=(6,0)$

$Hope~this~helps~and,\\Best~of~luck!\\\\~~~~~-TotallyNotTrillex$

"ටᆼට"

8 0

4 years ago

The length of a rectangle is 3 units shorter than one-third of the width, x.

AveGali [126]

$\frac{8x}{3} - 6$ is the expression that represents the perimeter of rectangle

Solution:

Let "x" represent the width of rectangle

Given that length of rectangle is 3 units shorter than one-third of the width x

length of rectangle = one-third of the width x - 3

$\text{ length of rectangle} = \frac{1}{3}x - 3 = \frac{x}{3} - 3\\\\\text{ length of rectangle} = \frac{x}{3} - 3$

The perimeter of rectangle is given as:

<h3>perimeter of rectangle = 2(length + width)</h3>

Substituting the known values we get,

$\text{ perimeter of rectangle }= 2(\frac{x}{3} - 3 + x)\\\\\rightarrow 2(\frac{x - 9 + 3x}{3})\\\\\rightarrow 2(\frac{4x-9}{3})\\\\\rightarrow\frac{8x-18}{3}\\\\\rightarrow\frac{8x}{3} - \frac{18}{3}\\\\\rightarrow\frac{8x}{3} - 6$

Thus perimeter of rectangle is $\frac{8x}{3} - 6$

Thus option 1 is correct

6 0

3 years ago