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inn [45]
3 years ago
12

Find the area of the composite figure.

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

25m²

Step-by-step explanation:

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Enlist your parents help to find you a decent tutor, it is much better to meet face to face.

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3 years ago
6. A Texas school district wants to compare the effectiveness of a standard AP Statistics curriculum and a new "hands-on" AP Sta
Butoxors [25]

The subjects in the experiment are given as follows:

b) the two AP Statistics curricula.

<h3>What are the subjects of an experiment?</h3>

The subjects of an experiment are the hypotheses which are studied in the experiment.

For this problem, we are testing hypotheses involving two forms of the curriculum, hence the correct option is given by:

b) the two AP Statistics curricula.

More can be learned about the subjects of an experiment at brainly.com/question/2792045

#SPJ1

5 0
2 years ago
In the following Laplace transform, how do you get rid of the cos(2at)*sin(at)?
Novay_Z [31]
<span>sin A cos B=1/2[sin(A-B)+sin(A+B)]
sin(at)*cos(2at)=1/2[sin(3at)-sin(at)]</span>
<span>
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7 0
3 years ago
Gabrielle is 3 times older than mikhail. the sum of their ages is 80. how old is mikhail
skelet666 [1.2K]
The sum of their ages equals 80.
Gabrielle is 3 times older than Mikhail

Two numbers that have a sum of 80, with one number being 3 times the other are:
Gabrielle’s age: 60
Mikhail’s age: 20

60 + 20 = 80
20 x 3 = 60


Mikhail is 20 years old.

4 0
3 years ago
Help please you don’t know how much this means to me
ICE Princess25 [194]
<h2>Problem 6:</h2><h2>a)</h2>

a(0) = 1  \:  \:  \:  \:  \:  \: b(0) = 2 \:  \:  \:  \:  \: c(0) = 3 \\ a(1) = b(0) + c(0) = 2 + 3 = 5 \\ b(1) = a(0) + c(0) = 1 + 3 = 4 \\ c(1) = a(0) + b(0) = 1 + 2 = 3 \\  \\ a(2) = b(1) + c(1) = 4 + 3 = 7 \\ b(2) = a(1) + c(1) = 5 + 3 = 8 \\ c(2) = a(1) + b(1) = 5 + 4 = 9

a(3) = b(2) + c(2) =  8 + 9 = 17\\ b(3) = a(2) + c(2) =7 + 9 = 16  \\ c(3) = a(2) + b(2) = 7 + 8 = 15 \\  \\ a(4) = b(3) + c(3) = 16  + 15 = 31 \\ b(4) = a(3) + c(3) = 17 + 15 = 32 \\ c(4) = a(3) + b(3) = 17 + 16 = 33

a(5) = 32 + 33 = 65 \\ b(5) = 31 + 33 = 64 \\ c(5) =31 + 32 =  63 \\  \\ a(6) = 64 + 63 = 127 \\ b(6) = 65 + 63 = 128 \\ c(6) = 65 + 64 = 129 \\

a(7) = 128 + 129 = 257 \\ b(7) = 127 + 129 = 256 \\c (7) = 127 + 128 = 255 \\  \\ a(8) = 256 + 255 = 511 \\ b(8) = 257 + 255 = 512 \\ c(8) = 257 + 256 = 513

a(9) = 512 + 513 = 1025 \\ b(9) = 511 + 513 = 1024 \\ c(9) = 511 + 512 = 1023 \\  \\ a(10) = 1024 + 1023 = 2047 \\ b(10) = 1025  + 1023 = 2048 \\ c(10) = 1025 + 1024 = 2049

<h2>b)</h2>

a(n) + b(n) + c(n) =   \\  2(a(n - 1) + b(n - 1) + c(n - 1)) \\   6 \times 2 {}^{n }

<h2>c)</h2>

6 \times 2 {}^{n}  > 100 \: 000 \\ 2 {}^{n}  >  \frac{100 \: 000}{6}  \\ n >  log {}^{2} ( \frac{100 \: 000}{6} )  \\ n > 14.02468 \\ n = 15

<h2 />
5 0
2 years ago
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