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ivolga24 [154]
3 years ago
7

2

Mathematics
1 answer:
Mama L [17]3 years ago
6 0

Answer:

od = 8 \sqrt{7}

Step-by-step explanation:

Apply pythagorean theorem.

eo {}^{2}  + od {}^{2}  = ed {}^{2}

EO is the radius so it is half of the diameter.

EO=24. ED is the hypotenuse.

24 {}^{2}  + od {}^{2}  = 32 {}^{2}

576 + od {}^{2}  = 1024

od {}^{2}  = 448

od = 8 \sqrt{7}

s

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σ = $3000, the maximum variance

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s = √σ = √(3000) = 54.772

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Answer:

240 cm², A = 300 cm², and 540 cm²

Step-by-step explanation:

1) The area of a parallelogram is A = bh, so:

A = (20)(12)

A = 240 cm²

2) Similarly, the area of a rectangle is A = bh, so:

A = (15)(20)

A = 300 cm²

3) The composite area is the sum of both the rectangle's area and the parallelogram's area, so:

240 cm² + A = 300 cm² = 540 cm²

3 0
3 years ago
Read 2 more answers
Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30
TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

-7, -3, 1, 5, . . .

An arithmetic sequence has a constant difference d and is defined by

a_n=a_1+\left(n-1\right)d

\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n

-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4

\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}

d=4

\mathrm{The\:first\:element\:of\:the\:sequence\:is}

a_1=-7

as

a_n=a_1+\left(n-1\right)d

\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:

a_n=4\left(n-1\right)-7           ∵ d=4

\mathrm{Arithmetic\:sequence\:sum\:formula:}

n\left(a_1+\frac{d\left(n-1\right)}{2}\right)

\mathrm{Plug\:in\:the\:values:}

n=30,\:\space a_1=-7,\:\spaced=4

=30\left(-7+\frac{4\left(30-1\right)}{2}\right)

=30\left(58-7\right)     ∵  \frac{4\left(30-1\right)}{2}=58

=30\cdot \:51

=1530           ∵  \mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

a_n=4\left(n-1\right)-7

n = 30

\sum _{n=1}^{30}\:4\left(n-1\right)-7

=\sum _{n=1}^{30}4n-11

\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n

=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11

as

\sum _{n=1}^{30}4n=1860

and

\sum _{n=1}^{30}11=330

so

=1860-330

=1530

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