Find the sum of the complex numbers

Quadrilateral ABCD is a parallelogram. Given this, fill in the coordinates below.

Lubov Fominskaja [6]

Answer:

The coordinates of D are (8,3)

Step-by-step explanation:

We have quadrilateral ABCD being a parallelogram with coordinates A(3,2), B(5,4), C(10,5), and D(a,b).

We want to find the coordinates of D.

Since ABCD is a parallelogram, the diagonals AC and BD bisects each other.

The midpoint of AC is (13/2,7/2).

This is also the midpoint of BD.

By the midpoint rule:

$( \frac{5 + a}{2} , \frac{4 + b}{2} ) = ( \frac{13}{2} , \frac{7}{2} )$

This implies that:

$5 + a = 13 \\ a = 13 - 5 \\ a = 8$

Similarly,

$4 + b = 7 \\ b = 7 - 4 \\ b = 3$

The coordinates of D are (8,3)

8 0

3 years ago

The number line is a graph of the _____.

vova2212 [387]

The number line is a graph of the Integers

6 0

4 years ago

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How do you convert 0.65 into a fraction

notsponge [240]

So its 0.65 so make that 65. then make it 65 over 100 then find 65 and 100's gcf (greatest common factor) then divide until its fully reduced

7 0

3 years ago

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X/x-2 + x-1/x+1= -1<br> Can someone walk me through the steps of this problem? Much appreciated. :)

stealth61 [152]

Answer:

x = 0 or x = 1

Step-by-step explanation:

Given equation:

$\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1$

Multiply the two denominators to get the same denominator:

Multiply the numerator of the first fraction by the denominator of the second fraction to get the new numerator of the first fraction.

Multiply the numerator of the second fraction by the denominator of the first fraction to get the new numerator of the second fraction.

$\implies \dfrac{x(x+1)}{(x-2)(x+1)}+\dfrac{(x-2)(x-1)}{(x-2)(x+1)}=-1$

Add the numerators and keep the denominator so that there is now one fraction:

$\implies \dfrac{x(x+1)+(x-2)(x-1)}{(x-2)(x+1)}=-1$

Simplify the numerator by expanding the brackets:

$\implies \dfrac{x^2+x+x^2-3x+2}{(x-2)(x+1)}=-1$

$\implies \dfrac{2x^2-2x+2}{(x-2)(x+1)}=-1$

Simplify the denominator by expanding the brackets:

$\implies \dfrac{2x^2-2x+2}{x^2-x-2}=-1$

Multiply both sides by the denominator of the left side:

$\implies 2x^2-2x+2=-1(x^2-x-2)$

Simplify and move everything to the left side:

$\implies 2x^2-2x+2=-x^2+x+2$

$\implies 3x^2-3x=0$

Factor:

$\implies 3x(x-1)=0$

Therefore:

$\implies 3x=0 \implies x=0$

$\implies x-1=0 \implies x=1$

Therefore, x = 0 or x = 1

8 0

2 years ago

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7.13 is 12.4% of what number?

Ksivusya [100]

7.13 is 12.4% of 57.5

8 0

4 years ago

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