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aliina [53]
3 years ago
7

TRANSFORMATI IN CENTIMETRI:20MM,60MM,300MM,670MM

Mathematics
2 answers:
svp [43]3 years ago
8 0
cm-centimeter\\mm-milimeter\\\\1cm=10mm\to1mm=0.1cm\\-----------------------\\20mm=20\cdot0.1cm=\boxed{2cm}\\60mm=60\cdot0.1cm=\boxed{6cm}\\300mm=300\cdot0.1cm=\boxed{30cm}\\670mm=670\cdot0.1cm=\boxed{67cm}
Vika [28.1K]3 years ago
3 0
Hey there! Every 10 millimeters (mm) is equal to 1 centimeter (cm), so we can simply divide each of these numbers by 10 to find the answers.
2 cm, 6 cm, 30 cm, and 67 cm.
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A circle has a diameter of 6, an area of b square units, and a circumference of c units. What is the value of b + c ?
marin [14]

Answer:

Step-by-step explanation:

Diameter = 6 units

Radius = Diameter/2 = 6/2 = 3 units

Taking π=3.14

Circumference = 2(π)(r) = 2(3.14)(3) = 6(3.14) = 18.84 = c

c = 18.84

Area = π(r)^2 = π(3)^2 = 9π =9(3.14) = 28.26 = b

b = 28.26

So,

b + c = 28.26 + 18.84 = 47.1

7 0
3 years ago
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likoan [24]

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8 0
2 years ago
The population of Greater Accra increased by 20% and became 900,7000. The original population of the city was:
allochka39001 [22]

Answer:

the original population of the city is 7,505,833

Step-by-step explanation:

Given;

new of the population of the city, Y = 9,007,000

the percentage increase of the city, = 20% = 0.2

Let the original population of the city = X

New population, Y = X  + 0.2X

                           Y = X(1 + 0.2)

                           Y = X(1.2)

                           Y = 1.2X

                           X = Y/1.2

                           X = (9,007,000) / 1.2

                           X = 7,505,833

Therefore the original population of the city is 7,505,833

3 0
3 years ago
Find the complete factored form of the polynomial -44a^3 + 20a^6
irakobra [83]

Answer:

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Step-by-step explanation:

→Take out the GCF (Greatest Common Factor). The GCF is 4a^3 because both -44a^3 and 20a^6 can be divided by it, like so:

-44a^3+20a^6

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3 years ago
Let U=[1,23,...,16] C=[1,3,5,...,15] Use the roster method to write the set C'.
Mazyrski [523]

The <em>complement </em>C'<em> </em>of some set C is essentially the set we get when we remove all of the elements of C from the universal set U.

Here, our universal set contains all of the numbers 1-16, and the set C contains all of the odd numbers between 1 and 16. If we were to put both of these sets in roster form:

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If we remove all of the elements of C from the universal set U, we're left with the complement of C, which contains all of the <em>even numbers </em>between 1 and 16:

C'=\{2,4,6,8,10,12,14,16\}

6 0
3 years ago
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