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maks197457 [2]
3 years ago
11

Okay so I’m stuck on 5. It says “ 10% of __ is 7” what’s the __ answer?

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

Answer:

70

Step-by-step explanation:

Okay so 10%=0.1. let's say that mystery number is x

that means that 0.1x=7

now to solve for x we need to multiply both sides of the equal sign by 10

so.....x=7*10=70        

Marat540 [252]3 years ago
3 0
10% of 70 is 7
.1 times 70 is 7
so the answer should be 70
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Answer:

Step-by-step explanation:

The question reads much more complicatedly than the actual equation.

Sakura * x = Nanuto

412 x = 634     Divide by 412

x = 634/412

x = 1.54 times

The closest answer is 3/2 as far. 1.54 is just about 3/2.

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What is the difference, in hours and minutes, between these times?
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Do the ratios 42/56 and 56/64 form a proportion?<br> if they do explain your answer
Volgvan
Lets check...

42/56 = 56/64
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(56)(56) = (42)(64)
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se necesita llenar un galon de agua,en 2 litros se tarda 30 minutos.cuantos litros me hacen falta para completar el galon y cual
mariarad [96]

Answer: 1.78 L are needed to complete the gallon and it will take 29.66 min to completely fill it

Step-by-step explanation:

The question in english is:

A gallon of water needs to be filled, in 2 liters it takes 30 minutes. How many liters do I need to complete the gallon and what is the time it takes me to fill it?

We know 1 gallon is equal to 3.78 liters:

1 gal=3.78 L

In addition, we are told you have filled the gallon with 2 L at a rate of \frac{2 L}{30 min}=0.06 L/min.

So, if we want to know how many liters are left to completely fill the gallon, we have to sbstract the 2 L to the total amount of a gallon:

3.78 L-2 L=1.78 L This is the amount you need to complete the gallon

On the other hand, the rate is defined as:

rate=\frac{Liters}{time}

Isolating time:

time=\frac{Liters}{rate}

Then, if we input the 1.78 L we need to fill the gallon we will have the time:

time=\frac{1.78 L}{0.06 L/min}

time=29.66 min This is the time you need to fill the gallon

6 0
3 years ago
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