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Pani-rosa [81]
3 years ago
13

Follow the steps to find the quotient. Please help !

Mathematics
2 answers:
kirza4 [7]3 years ago
7 0

Answer:

Nothing goes into 3

8 goes into 32  four times (8 times 4= 32) so that means you'd put 4 on top of 2

then you bring down the 8 and 8 goes into 8 once. (8 times 1= 8) so you'd put 1 on top of 8.

Now down below you'd put 32 below 32 since you subtract it.

I'm not sure what the 3 bottom ones are.

I think you'd put 8 in the first  boxes then 0 cause 8=8 = 0.

good luck <3

answer is 41 <3

ale4655 [162]3 years ago
6 0

Answer:

41

Step-by-step explanation:

1. Look at the first digit and ask yourself: is this divisible by 8? (3 is not so you look at the next one)

2. 8 goes into 32 evenly 4 times (put 4 at the top)

3. Subtract 42

4. Drag down the 8

5. Does 8 go into 8? (Yes!)

6. Put one at the top and subtract to get zero

7. Answer is at the top: 41

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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.

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<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
4 years ago
The population of a city on April 15, 1915, was 47,175. During the period between March 1 and July 1, 1915, 1,325 new cases of s
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Answer:

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Step-by-step explanation:

The computation of the monthly incident rate of active cases is shown below:

Incidence rate is

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Which equation is equivalent
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The equation that is equivalent is -2+(-7).
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