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OLga [1]
3 years ago
6

What is the decimal equivalent to the fraction 80/32

Mathematics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

2.5

Step-by-step explanation:

\frac{80 \div 16}{32 \div 16}  =  \frac{5}{2}

first you have to decide by a number that goes into both the numator and denominator in this case you dive the top and bottom by 32. witch equals

\frac{5}{2}

then you divide 5 and 2

5 \div 2

which equals

5 \div 2 = 2.5

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Use
Brums [2.3K]

Answer:

I don't know what inductive reasoning is but, the next two numbers in the pattern are -243 and +729

Step-by-step explanation:

each number is being multiplied by a factor of (-3).

8 0
2 years ago
the width of a rectangle is 2/3 length, The perimeter of the rectangle is 100 ft. What is the width, in feet, of the rectangle.
lakkis [162]

Answer:

The width would be 20 and the length would be 30

Step-by-step explanation:

7 0
3 years ago
Calculate the length of the line segment round to the nearest whole number 6,6 and -2,-3
Nataliya [291]
It is 14 because 8= 6to-2 and 3to-3is 6 so both together is 14
hope this helps
4 0
3 years ago
Ahsan
NikAS [45]

It is number 1 and number 2 because the equation is 242× 62 and Number 1 and 2 are just breaking it down

7 0
3 years ago
Approximate the change in the volume of a sphere when its radius changes from r​ = 40 ft to r equals 40.05 ft (Upper V (r )equal
alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere (V), measured in cubic feet, is represented by the following formula:

V = \frac{4\pi}{3}\cdot r^{3}

Where r is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

\Delta V = 4\pi \cdot r^{2}\cdot \Delta r

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0
3 years ago
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