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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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Kamau bought 50 kg of sugar for sh 4000
Nana76 [90]

Answer:

Profit % =50%

Step-by-step explanation:

Cost of 50 kg = sh 4000

Selling price of 1kg = sh 120

Selling price of 50 kg = sh 6000

profit \% = \frac{Selling \ price - cost \ price}{cost \ price} \times 100\\

            =\frac{6000 - 4000}{4000} \times 100\\\\=\frac{2000}{4000} \times 100\\\\=\frac{1}{2} \times 100\\\\=50 \%

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Answer:

Diameter is 19

Step-by-step explanation:

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3 years ago
How many significant figures will there be in the answer to the following problem? you do not have to solve the problem. 223.4 •
Irina-Kira [14]

Answer:

5 significant figures

Step-by-step explanation:

looking at the last two digit of the numbers, it multiplication gives, 4*5=20

we already have four significant figures ,

so, 2 would be the fifth since 0 is not going to be counted

3 0
3 years ago
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vampirchik [111]
The answer is 13 hope I helped
5 0
3 years ago
Read 2 more answers
4. Diego is thinking of two positive numbers. He says, "If we triple the first number and
aleksandr82 [10.1K]

Answer:

3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

Now,  3<em>x </em>+ 2<em>y</em> = 34

3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

30 = 34 - 2y

2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

2y = 22

y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

8 0
3 years ago
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