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Zolol [24]
3 years ago
12

a quadrilateral has vertices A(11,-7), B(9,-4), C(11,-1), and D(13,4). Quadrilateral ABCD is a . If the vertix C(11,-1) were shi

fted to the point C(11,1), quadrilateral ABCD would be a .
Mathematics
2 answers:
Maurinko [17]3 years ago
7 0

Answer:

A quadrilateral formed by the vertices A(11,-7), B(9,-4), C(11,-1), and D(13,4) forms a rhombus

If the vertex C(11,-1) were shifted to the point C'(11,1), quadrilateral ABC'D would be a kite

Step-by-step explanation:

A quadrilateral formed by the vertices A(11,-7), B(9,-4), C(11,-1), and D(13,4) forms a rhombus. A rhombus is a quadrilateral whose all sides have equal length means it has four equal sides.

If the vertex C(11,-1) were shifted to the point C'(11,1), quadrilateral ABC'D would be a kite because a kite has two pairs of equal sides. A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides which are adjacent to each other....

postnew [5]3 years ago
5 0

Answer:

a. rhombus b. kite

Step-by-step explanation:

plato

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Simple linear equation is :

Y = 86.784 - 2.67x

Step-by-step explanation:

The educator wants to see the relationship between no of absences and final grades of the student. No. of absence of student is independent variable while final grade is dependent variable. The final grades of the student are dependent on the no. of absence they do.

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b =  { 2312 [ 37 * 422 ] / 6 } / 337 - 37^2 / 6 = -2.67

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Describe the relationship between the point D(6, 9) and the point A(8, 12) in terms of dilations.
Irina18 [472]

Answer:

one is smaller then the other i think thats it

Step-by-step explanation:

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2 years ago
More 5th grade work c’mon
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Answer: honestly for the first one, i think the answer would be 2.

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Step-by-step explanation:

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Step-by-step explanation:

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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
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3 years ago
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