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Dafna1 [17]
3 years ago
15

I need help on this question can someone please help me?

Mathematics
2 answers:
just olya [345]3 years ago
5 0
Opposite angles in a parallelogram are equal so
4n - 2 = 2n + 32
you move the 2n to the left side of the = (trying to get all the n values on one side of the equation) and move the - 2 to the right side of the =
when you move a value to the opposite side of the = the thing becomes the opposite. so 2n is a positive so when you move it to the left it becomes a negative so -2n. the -2 is a minus so when you move it to the right side of the = it becomes a plus so +2
4n - 2n = 32 + 2
you add/subtract the like terms so
2n = 34
then you move the 2 from the 2n to the opposite side of the = as you’re trying to get n on its own. 2n has a hidden multiply (2 x n) so when you move it to the opposite side of the = the multiply becomes the opposite which is divide so the equation is now
n = 34 ÷ 2
n = 17
Svetllana [295]3 years ago
3 0

I really hate these algebra problems pretending to be geometry.  

Anyway angles B and D  are congruent in a parallelogram.  Geometry part done.

4n - 2 = 2n + 32

2n = 34

n = 17

Answer: 17, which is the 3rd choice

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<u><em>6% decline</em></u>

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1. Trapezoid KLMN has vertices K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2) .
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Answer:


Step-by-step explanation:

(A) The vertices of the trapezoid KLMN are K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2)

Now, KL= \sqrt{(3-1)^{2}+(1-3)^{2}}=\sqrt{8}[/tex]

LM=\sqrt{(3-3)^{2}+(0-1)^{2}}=1

MN=\sqrt{(1-3)^{2}+(-2-0)^{2}}=\sqrt{8}

NK=\sqrt{(1-1)^{2}+(-2-3)^{2}}=5

Now, as KL and MN  are equal, therefore, KLMN is an isosceles trapezoid.

(B) Since  m∠XWY=47°, therefore ∠XWZ=47+47=94° and ∠ZYW=18°, therefore ∠XYZ=36°( as diagonals bisect the angles)

In ΔXWO,

∠XWO+∠WXO+∠WOX=180°(Angle sum property)

∠WXO=43°

Also, from ΔXOY,

∠OXY=72° using the angle sum property.

Therefore, ∠WXY=43+72=115°

Now, sum of all the angles of a quadrilateral is equal to 360°, therefore

∠WXY+∠XYZ+∠YZW+∠ZWX=360°

115°+36°+∠YZW+94°=360°

∠YZW=115°

Therefore,∠WZY=115°

(C) Since, AB and CD are the two parallel lines as the slope of both the sides are equal.

LetM be the mid point of AB, therefore M=(\frac{-2+4}{2}, \frac{4+3}{2})

=(1,\frac{7}{2})

Also, let N be the mid point of DC, therefore,

N=(\frac{4-2}{2},\frac{-2-5}{2})

=(1,\frac{-7}{2})

Now, length of the mid segment MN= \sqrt{(1-1)^{2}+(\frac{7}{2}+\frac{7}{2})^{2}}=\sqrt{0+49}=7 cm

(D)Given: Kite PQRS,  TS=6cm and TP=8cm

Then, From triangle TSP, we have

(SP)^{2}=(TS)^{2}+(TP)^{2}

(SP)^{2}=36+64

SP=10cm

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