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elena-s [515]
3 years ago
9

Pls help it urgent pls empathise

Mathematics
1 answer:
IRINA_888 [86]3 years ago
7 0
#17
Use property that the diagonals of a parallelogram bisect each other, =>
x+y=18
 y+(can't read)=22
Solve for x and y.

#18
z+60=180   [angles on a line are supplementary]
x=z   [opposite angles of a parallelogram are congruent]
y=50 [alternate interior angles between parallel lines AD and BC]

#19
3v=30  [ opposite sides of a parallelogram are congruent ]
solve for v.
2u-1=19 [ ditto ]
solve for u.

#20
(a) AB//DC [ adjacent angles are supplementary ]
(b) PS//QR [ ditto ]

#21
1) a parallelogram with one angle 90 means the adjacent angles are 180-90=90.  This in turn means all angles are 90, so the figure is a rectangle (adjacent sides may or may not be equal in a parallelogram).
2) a rectangle with adjacent sides equal must be a square.

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Pls help me i need to do all my math for to day and i dont know how to do it <br><br> -11x(x+12)
Vesna [10]

Answer:

If you want to simplify, -11x^{2}-132x

Step-by-step explanation:

4 0
3 years ago
Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f
GalinKa [24]

Answer:

\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in \mathbb{R}.

\\f(x)=(x^2+16)(x^2-2x-15)

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}

And we can write in \mathbb{C}

f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago
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snow_tiger [21]

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5 0
4 years ago
Read 2 more answers
If the solution to a linear system of equations is x=6 and y= 2 and you graphed the two equations, where would the two lines int
coldgirl [10]
<h3>The lines intersect at (x,y) = (6,2)</h3>

This is simply because the x coordinate is 6 and the y coordinate is 2.

The line x = 6 is a vertical line in which all points on it have an x coordinate of 6. It goes through 6 on the x axis.

The line y = 2 is horizontal where all points on it have a y coordinate of 2.

The vertical and horizontal lines intersect at (6,2)

8 0
2 years ago
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