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ExtremeBDS [4]
4 years ago
13

Determine the precise name for quadrilateral WXYZ

Mathematics
2 answers:
Archy [21]4 years ago
7 0
It is a trapezoid so the first one
Natali5045456 [20]4 years ago
3 0

it is a quadrilateral

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the providence children muesuem installed a fence around the perimeter of the propety. the propety is a rectangle. one side is 3
mash [69]
Its a rectangle so two sides are equal and both sides that are opposite of each other are equal. So two sides are 37. Then you add them together to get 74, then subtract it from 164 to get 90 the rest of the fence length. Then divide it by two to get 45 which is the other side lengths, so it would be a 37 by 45 rectangle one side would be 37 while the other two are 45.
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3 years ago
How many 1-centimeter squares fit along the length of the rectangle
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It would depend on how large the rectangle is
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3 years ago
What are the steps used to transform the general form of the equation of a circle to standard form
Bezzdna [24]

Answer:

Step-by-step explanation:

Convert the equation of a circle in general form shown below into standard form. Find the center and radius of the circle. Group the x 's and y 's together. Consider the x2 and x terms only. Complete the square on these terms. Replace the x2 and x terms with a squared bracket.

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3 years ago
Show 16300000 in standard form
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7 0
3 years ago
How to solve this :') please help
blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

so

\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}

7 0
2 years ago
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