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bazaltina [42]
1 year ago
12

How would I do fractions for common difference for each arithmetic sequence?

Mathematics
1 answer:
Sedbober [7]1 year ago
6 0

In an arithmetic sequence, you add or subtract a fixed amount called the common difference to be able to get the next term in the sequence.

The first step in solving an arithmetic sequence containing fractions, you subtract the first term from the second term to get the common difference.Then using this common difference you find the nth term asked.

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Ming has 15 quarters, 30 dimes, and 48 nickels. He wants to group his money so that each group had the same number of each coin.
Colt1911 [192]
Find the greatest common factor. In this case it is 3. Divide each group by 3.
15÷3=5 quarters
30÷3=10 dimes
48÷3=16 nickels
Next count them by the coin's value:
5 × 0.25 = $1.25
10 × 0.10 = $1
16 × 0.05 = $0.80
Add them up:
= $3.05 in each group and there are three groups.

So, the greatest number of groups that he can make is 3; there will be 5 quarters, 10 dimes, and 16 nickels in each group, which is worth $3.05 in each group.
3 0
3 years ago
Find x in this 45°-45°-90° triangle.<br> x=
vova2212 [387]

Answer:

33

Step-by-step explanation:

because a triangle measure 90 degrees so you just subtract the amount that you add and got

6 0
3 years ago
Judy can decorate 3 cakes in 5 hours which graph has the slope that best represents the number of cakes per hour Judy can decora
aniked [119]

Answer:

The slope = 3 / 5

Step-by-step explanation:

Image attached shows the relationship between cakes and time

6 0
3 years ago
Given A(-1,4) B(1,5) and C(-5,3) which coordinate will make AB parallel to CD?
Solnce55 [7]
I think b not sure tho
7 0
3 years ago
Not sure how to do #69, it's a calc 1 question
Gnoma [55]
Let A( t , f( t ) ) be the point(s) at which the graph of the function has a horizontal tangent => f ' ( t ) = 0.

But, f ' ( x ) = [ ( x^2 ) ' * ( x - 1 ) - ( x^2 ) * ( x - 1 )' ] / ( x - 1 )^2 =>
f ' ( x ) = [ 2x( x - 1 ) - ( x^2 ) * 1 ] / ( x - 1 )^2 => f ' ( x ) = ( x^2 - 2x ) / ( x - 1 )^2;

f ' ( t ) = 0 <=> t^2 - 2t = 0 <=> t * ( t - 2 ) = 0 <=> t = 0 or t = 2 => f ( 0 ) = 0; f ( 2 ) = 4 => A 1 ( 0 , 0 ) and A 2 ( 2 , 4 ).
3 0
3 years ago
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