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d1i1m1o1n [39]
2 years ago
12

How would the factor pairs change if nancy had only 5 posters to arrange?

Mathematics
1 answer:
Romashka [77]2 years ago
3 0

Answer:

Part A) Nancy can arrange the posters in 4 ways

Part B) Nancy can arrange the posters in 2 ways

Step-by-step explanation:

<u><em>The complete question is</em></u>

Nancy has 10 movie posters. She wants to hang them on a wall in equals rows.

A) Find all the ways that she can arrange the posters.

B) How would the factor pairs change if Nancy had only 5 posters to arrange?

Part A) Find all the ways that she can arrange the posters.

we know that

The factors of 10 are

1x10 -----> 1 row of 10 posters

2x5 -----> 2 rows of 5 posters

5x2 -----> 5 rows of 2 posters

10x1 -----> 10 rows of 1 poster

therefore

Nancy can arrange the posters in 4 ways

Part B) How would the factor pairs change if Nancy had only 5 posters to arrange?

we know that

if Nancy had only 5 posters to arrange

then

The factors of 5 are

1x5 -----> 1 row of 5 posters

5x1 -----> 5 rows of 1 poster

therefore

Nancy can arrange the posters in 2 ways

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What is the rate of change of the function?
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3

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Which value is equivalent to: 1 1/2 divided by 4 1/2 x 1 1/3
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Step-by-step explanation:

Hi. Here's the work.

First turn everything into an improper fraction.

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6 0
2 years ago
Pls help anyone us appreciated
MAVERICK [17]

x=angle CDB  (vertical angles)

=angle ABE (corresponding angles, AB parallel to CD)

=132-90   (exterior angle 132=sum of angles ABE and AEB)

=42 degrees

5 0
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3) Suppose you deposit $10,000 in a savings account that pays interest at an annual rate of 5%
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9514 1404 393

Answer:

  3.67 years

Step-by-step explanation:

The amount is found using the compound interest formula.

  A = P(1 +r/n)^(nt)

for principal P invested at annual rate r compounded n times per year for t years.

We can solve this for t:

  A/P = (1 +r/n)^(nt) . . . . divide by P

  log(A/P) = (nt)log(1 +r/n) . . . . take the logarithm

  t = log(A/P)/(n·log(1 +r/n)) . . . . divide by the coefficient of t

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