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3 years ago

9

Can someone please help me with this I literally have no idea what I'm doing. Thanks!

1 answer:

Phoenix [80]3 years ago

6 0

The point that divides the line from M to E in ratio 1:3 is (-7/2,8)

Step-by-step explanation:

Given points are:

M(-6,7) = (x1,y1)

E(4,11) = (x2,y2)

Let the ratio be a:b

Then from 1:3

a=1

b=3

The formula for finding the point that divides the line segment into ratio a:b is:

$(\frac{bx_1+ax_2}{a+b},\frac{by_1+ay_2}{a+b})\\Putting\ the\ values\\(\frac{(3)(-6)+(1)(4)}{1+3},\frac{(3)(7)+(1)(11)}{1+3})\\(\frac{-18+4}{4},\frac{21+11}{4})\\(\frac{-14}{4},\frac{32}{4})\\(\frac{-7}{2},8)$

Hence,

The point that divides the line from M to E in ratio 1:3 is (-7/2,8)

Keywords: Coordinate Geometry, Lines

Learn more about coordinate geometry at:

#LearnwithBrainly

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g The fencing of the left border costs $10 per foot, while the fencing of the lower border costs $2 per foot. (No fencing is req

klio [65]

Answer:

$Area = 1690ft^2\\x = 130ft, y = 26ft$

Step-by-step explanation:

Given the costs we can form an equation:

$10y + 2x =520 - Eq(A)$

and fencing is triangular such that the area enclosed can be written as:

$A = \dfrac{xy}{2} -Eq(B)$

First need to convert the above equation so that it is only in terms of one variable. [either x or y]

To make the equation only in terms of $x$ we can substitute $y$ from Eq(A) i.e, $y =\frac{520-2x}{10}$ , to Eq(B)

$A = \dfrac{x}{2} \dfrac{520-2x}{10}$

simplify

$A = \dfrac{520x - 2x^2}{20}$

$A = -\dfrac{1}{10}x^2 + 26x$

Now, in order to find the maximum area enclosed we can find $\frac{dA}{dx}$ and equate it zero.

$\dfrac{dA}{dx} = -\dfrac{1}{5}x + 26$

$0 = -\dfrac{1}{5}x + 26$

$-26 = -\dfrac{1}{5}x$

$x = 130$

we have the length of one dimension: specifically, the lower fence will be $x =130ft$

we can use this value of $x$ to find the corresponding value of $y$ . From Eq(A)

$10y + 2x =520$

$10y +2(130) = 520$

$y = \dfrac{520 - 2(130)}{10}$

$y = 26$

the length of the left fence will be $y =26ft$

The enclosed area by the fence will be

$A = \dfrac{xy}{2}$

$A = \dfrac{(130)(26)}{2}$

$A = 1690$

Hence the maximum area that can enclosed by the fences provided the costs will be $1690ft^2$

You can even check the cost of the dimensions whether they all add up to $520 or not.

Use Eq(A)

$10y + 2x =520$

$10(26) + 2(130) =520$

and indeed it does!

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