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padilas [110]
3 years ago
6

5 Exam-style ABCD is a kite.

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
5 0

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}

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Arisa [49]

Answer:

It took 17 days for the pasture to be one-third covered in grass

Step-by-step explanation:

* Lets explain how to solve the problem

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∵ The number of blades of grass is triple every day

∴ The next day the number of blades of grass will be 3x

∴ The formula to find the number of blades of grass after d days is

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- After 18 days, the entire pasture is completely covered in grass

∵ d = 18

∴ The number of blades of grass = x(3)^{18}

- We need to know the the number of days that took for the pasture

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∵ The number of blades of grass for 1/3 covered = (\frac{1}{3})x(3)^{18}

∴ (\frac{1}{3})x(3)^{18}=x(3)^{d}

- Remember  \frac{1}{3}=3^{-1}

∴ 3^{-1}x(3)^{18}=x(3)^{d}

- Remember we add the power of the same bases when we

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∴ x(3)^{-1+18}=x(3)^{d} ⇒ divide both sides by x

∴ (3)^{17}=(3)^{d}

- Remember if the two bases equal in the equation, then their

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∴ d = 17

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4 0
3 years ago
Two models of cellular telephones, red and blue, are stored in boxes. One box weighs twelve pounds and contains four of the red
GenaCL600 [577]
First, make a equation in which r= red and b=blue.

So, since in the first box it has 4 red models and one blue model equaling 12,

the first equation looks like 4r+b=12.

The second equation looks like 2r+b=8.

What you would do is try and first solve for r by getting rid of b.

Since both equation has a positive b, you would make one equation have a negative b by multiplying the whole equation by -1.

-1*(2r+b=8)= -2r-b=-8

Add.

4r+b=12
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2r=4
r=2

Then, you plug in 2 for the r for any of the original equations.

4(2)+b=12

8+b=12

b=4

or

2(2)+b=8

4+b=8
b=4

So, the red models weigh 2 pounds while the blue models weigh 4.



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3 years ago
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ZanzabumX [31]

Note the equal sign. What you do to one side, you do to the other.


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8 0
3 years ago
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