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algol [13]
3 years ago
7

Prove that the diagonals of a parallelogram bisect each other. The midpoints are the same point, so the diagonals _____

Mathematics
1 answer:
Molodets [167]3 years ago
8 0

Answer:

Below

Step-by-step explanation:

To prove that the diagonals bisect each other we should prove that they have a common point.

From the graph we notice that this point is E.

ABCD is a paralellogram, so E is the midpoint of both diagonals.

●●●●●●●●●●●●●●●●●●●●●●●●

Let's start with AC.

● A(0,0)

● C(2a+2b,2c)

● E( (2a+2b+0)/2 , (2c+0)/2)

● E ( a+b, c)

●●●●●●●●●●●●●●●●●●●●●●●●

BD:

● B(2b,2c)

● D(2a,0)

● E ( (2a+2b)/2 , 2c/2)

● E ( a+b ,c)

●●●●●●●●●●●●●●●●●●●●●●●●●

So we conclude that the diagonals bisect each others in E.

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dalvyx [7]

Answer:

\boxed{\sf \ \ \ m = -\dfrac{tn}{t-s} \ \ \ }

Step-by-step explanation:

Hello,

let s assume that m+n is different from 0

we have this equation and we need to find m as a function of t, s, and n

t=\dfrac{ms}{m+n}

<=>

(m+n)*t=ms\\\\ tm+tn=sm\\ (t-s)m = -tn\\ m = -\dfrac{tn}{t-s}

for t-s different from 0, so t different from s

hope this helps

4 0
3 years ago
Jared's Cafe uses 5 bags of coffee every day. How many days will 1/2 of a bag of coffee last?
Inessa05 [86]

Answer:

kgyutguyguky

Step-by-step explanation:

4 0
2 years ago
If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha
castortr0y [4]
6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-\frac{2}{3}\\\\2x-1=0\to x=\frac{1}{2}\\\\\alpha=-\frac{2}{3};\ \beta=\frac{1}{2}\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{-\frac{2}{3}}{\frac{1}{2}}+\frac{\frac{1}{2}}{-\frac{2}{3}}=-\frac{2}{3}\cdot\frac{2}{1}-\frac{1}{2}\cdot\frac{3}{2}=-\frac{4}{3}-\frac{3}{4}\\\\=-\frac{16}{12}-\frac{9}{12}=-\frac{25}{12}=-2\frac{1}{12}


use\ Vieta's\ formula:\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2}{\alpha\beta}-2\\\\\alpha+\beta=\frac{-b}{a};\ \alpha\beta=\frac{c}{a}\\\\\frac{(\alpha+\beta)^2}{\alpha\beta}-2=\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-2=\frac{b^2}{a^2}\cdot\frac{a}{c}-2=\frac{b^2}{ac}-2

6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{1^2}{6\cdot(-2)}-2=\frac{1}{-12}-2=-2\frac{1}{12}
7 0
3 years ago
Hence find the value of x that satisfies the two equations y =
sammy [17]

Answer:

x=6

Step-by-step explanation:

since y=5x/2-7, and y=4x/3

         so 5x/2-7=4x/3

              5x/2-4x/3= 7

              15x-8x/6=7

               7x/6=7

               x= 7*6/7= 6

5 0
2 years ago
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