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AlekseyPX
3 years ago
9

Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-6, -4),A(−6

,−4),A, left parenthesis, minus, 6, comma, minus, 4, right parenthesis, comma B(-4,-4)B(−4,−4)B, left parenthesis, minus, 4, comma, minus, 4, right parenthesis, C(-4, -2)C(−4,−2)C, left parenthesis, minus, 4, comma, minus, 2, right parenthesis, and D(-6, -2)D(−6,−2)D, left parenthesis, minus, 6, comma, minus, 2, right parenthesis.
What is the perimeter of rectangle ABCDABCDA, B, C, D?

units

Mathematics
1 answer:
Lelu [443]3 years ago
8 0

Answer:

8

Step-by-step explanation:

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at a particular school 40% of children travel by bus if this represents 400 children, how many children attend the school
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560 students

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The principal P is borrowed at a simple interest rate are for a period of time T. Find the loans future value A, or the total am
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The total amount due after five years is $57,000.

Step-by-step explanation:

Recall that simple interest is given by the formula:

\displaystyle A=P(1+rt)

Where <em>A</em> is the final amount, <em>P</em> is the principal amount, <em>r</em> is the rate, and <em>t</em> is the time (in years).

Since we are investing a principal amount of $38,000 at a rate of 10.0% for five years, <em>P</em> = 38000, <em>r</em> = 0.1, and <em>t</em> = 5. Substitute:

\displaystyle A=38000(1+(0.1)(5))

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4 0
2 years ago
a hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. the c
allochka39001 [22]

According to given conditions, m+n is equal to 409.

Consider the diagram below.

In the hexagon ABCDEF, let

AB = BC = CD = 3;

DE = EF = FA = 5;

Arc BAF is equal to one-third of the circle's circumference.

Hence, ∠BCF = ∠BEF = 60°;

Similarly, ∠CBE = ∠CFE = 60°;

Let the point of intersection of BE and CF be P, BE and AD be Q and CF and AD be R.

∴ Δ EFP and Δ BCP are equilateral, and so Δ PQR is also equilateral.

Also, ∠ BAD and ∠ BED subtend the same arc and so do ∠ ABE and ∠ ADE.

∴ Δ ABQ is similar to Δ EDQ

\frac{AQ}{EQ}  = \frac{BQ}{DQ} = \frac{AB}{ED} = \frac{3}{5}

Also,

\frac{\frac{AD - PQ}{2} }{PQ + 5} = \frac{3}{5}  

and \frac{3 - PQ}{\frac{AD + PQ}{2} }  = \frac{3}{5}

On solving these simultaneous equations, we get AD = 360/49

∴ m + n = 409.

To learn more about similarity of triangles, refer to this link:

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