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emmasim [6.3K]
3 years ago
11

Write the cross products for the proportion s/t= u/v

Mathematics
1 answer:
Brrunno [24]3 years ago
8 0

Hello from MrBillDoesMath!

Answer:

sv = tu


Discussion:

For the case of a proportion the "cross product" refers to the products of the "extremes" equaling the products of the "means".  As (s,v) are the extremes and (u.t) the means, this gives sv = tu


 

Thank you,

MrB

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Answer:

P(t) = 282.2(1.009)^t

Step-by-step explanation:

Look at the attached image.

Hope you can read my handwriting. the image cut off the right side, b = 1.009213324... but question asks to round to nearest thousandth so it's 1.009

For the second part just use the equation to find P when t = -1 and see if P is less than (underpredicts) the number from the question or greater than (overpredicts) the number. I haven't calculated it but I think it will be smaller and thus underpredicts just from looking at the numbers when t = -1

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Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by \sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}.

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

a_{n} = a_{1} • r^{n-1}

Where a_{n} is the nth term, a_{1} is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, a_{1} will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

\sqrt[5]{\frac{a_{6}}{a_{1}}} =

\sqrt[5]{\frac{1}{32}} =

\frac{1}{2}.

Therefore: r = \frac{1}{2}.

Using all the information we have, we can find the explicit rule:

a_{n} = a_{1} • r^{n-1}

  • a_{1} = 32
  • r = \frac{1}{2}

a_{n} = a_{1} • r^{n-1} →

\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

a_{1} = 32 • (\frac{1}{2})^{1-1}

a_{1} = 32 • (\frac{1}{2})^{0}

a_{1} = 32 • 1

a_{1} = 32

a_{6} = 32 • (\frac{1}{2})^{6-1}

a_{6} = 32 • (\frac{1}{2})^{5}

a_{6} = 32 • \frac{1}{32}

a_{6} = 1

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