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Fed [463]
3 years ago
15

Help me plzz!!

Mathematics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

41/6 lb

Step-by-step explanation:

Start with the original problem

5 11/6

1.) Learn the parts of the mixed number. In this case, 5 is a whole number, 11 is a numerator, and the 6 is the denominator.

2.) to turn a mixed number into an improper fraction or a whole number, multiply the denominator times the whole number (6 * 5 = 30) then add the numerator to the value you got (30 + 11 = 41)

3.) If the improper fraction can be simplified then you will most likely need to go ahead and simplify it. In this case, 6 doesn't factor into 41 so you would keep it as 41/6.


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Geometric Sequences Question.
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2 years ago
What is the common ratio of the geometric sequence below? –2, 4, –8, 16, –32, ...
SOVA2 [1]

Answer:

The common ratio of the geometric sequence is:

r=-2

Step-by-step explanation:

A geometric sequence has a constant ratio 'r' and is defined by

a_n=a_1\cdot r^{n-1}

where

  • aₙ is the nth term
  • a₁ is the first term
  • r is the common ratio

Given the sequence

-2,\:4,-8,\:16,\:-32,\:...

Compute the ratios of all the adjacent terms:  r=\frac{a_n+1}{a_n}

\frac{4}{-2}=-2,\:\quad \frac{-8}{4}=-2,\:\quad \frac{16}{-8}=-2,\:\quad \frac{-32}{16}=-2

The ratio of all the adjacent terms is the same and equal to

r=-2

Therefore, the common ratio of the geometric sequence is:

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5 0
2 years ago
A metalworker has a metal alloy that is 20​% copper and another alloy that is 60​% copper. How many kilograms of each alloy shou
puteri [66]
<h3>Answer:</h3>
  • <u>20</u> kg of 20%
  • <u>80</u> kg of 60%
<h3>Step-by-step explanation:</h3>

I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.

That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.

_____

<em>Using an equation</em>

If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...

... 0.60x + 0.20(100 -x) = 0.52·100

... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20

... x = 32/0.40 = 80 . . . . . kg of 60% alloy

... (100 -80) = 20 . . . . . . . .kg of 20% alloy

6 0
3 years ago
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