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SashulF [63]
4 years ago
9

Whats 0.2 divide by 0.4 work and answer

Mathematics
2 answers:
alexdok [17]4 years ago
3 0
Its is two since you take out the decimals then divide regularly
inn [45]4 years ago
3 0
The answer for 0.2 divide by 0.4 is 0.5. I hope this helps.
You might be interested in
Plase dont awnser just for points
egoroff_w [7]
So 3 2/3 is about 3.67 so it would go in between 3.65 and 3.7. So it would be the second answer.
5 0
4 years ago
Read 2 more answers
What is the area of the kite?
Kazeer [188]

Multiply 4.93m by 8.5m to get 41.905m to the second power.

You do this because there are 2 identical triangles on the top. And if you put those two triangles together, you get a rectangle. The length of the rectangle is 8.5m while the width would be 4.93. Multiplying the length and width gives you the area.

Then multiply 10.2m by 8.5m to get 86.7m to the second power.

You do this because there are 2 identical triangles on the bottom. And if you put those two triangles together, you get another rectangle. The length of the rectangle is 8.5m while the width would be 10.2m. Multiplying those together gives you the area.

You then add the two areas, 41.905m to the second power and 86.7m to the second power, to get the area of the entire figure.

After adding, you get 128.605 m to the second power. That's the answer

5 0
3 years ago
Simplify 25/26•13/15<br> HELPME
leva [86]
\frac{25}{26} * \frac{13}{15} =  \frac{25~*~13}{26~*~15} =  \frac{325}{390} = \frac{5}{6}
6 0
4 years ago
Read 2 more answers
One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second
algol13

Answer:

Therefore the third container contains   \frac{735}{17}\% = 43.23 %  acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

= (V\times \frac{100+55}{100})   cubic unit.

=\frac{155}{100}V  cubic unit.

The amount of acid in first container is

=V\times \frac{25}{100}  cubic unit.

=\frac{25}{100}V  cubic unit.

The amount of acid in second container is

=(\frac{155}{100}V \times \frac{55}{100}) cubic unit.

=\frac{8525}{10000}V cubic unit.

Total amount of acid =(\frac{25}{100}V+\frac{8525}{10000}V) cubic unit.

                                   =(\frac{2500+8525}{10000}V) cubic unit.

                                   =(\frac{11025}{10000}V)cubic unit.

Total volume of mixture =(V+\frac{155}{100}V) cubic unit.

                                       =\frac{100+155}{100}V cubic unit.

                                       =\frac{255}{100}V  cubic unit.

The amount of acid in the mixture is

=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%

=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%

=\frac{735}{17}\%

Therefore the third container contains  \frac{735}{17}\%  acid.

5 0
3 years ago
5. The recursive algorithm given below can be used to compute gcd(a, b) where a and b are non-negative integer, not both zero.
s2008m [1.1K]

Implementating the given algorithm in python 3, the greatest common divisors of <em>(</em><em>124</em><em> </em><em>and</em><em> </em><em>244</em><em>)</em><em> </em>and <em>(</em><em>4424</em><em> </em><em>and</em><em> </em><em>2111</em><em>)</em><em> </em>are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

<em>#initialize</em><em> </em><em>a</em><em> </em><em>function</em><em> </em><em>named</em><em> </em><em>gcd</em><em> </em><em>which</em><em> </em><em>takes</em><em> </em><em>in</em><em> </em><em>two</em><em> </em><em>parameters</em><em> </em>

if a>b:

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>b</em>

return gcd (b, a)

<em>#if</em><em> </em><em>true</em><em> </em><em>interchange</em><em> </em><em>the</em><em> </em><em>Parameters</em><em> </em><em>and</em><em> </em><em>Recall</em><em> </em><em>the</em><em> </em><em>function</em><em> </em>

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

<em>#even</em><em> </em><em>numbers</em><em> </em><em>leave</em><em> </em><em>no</em><em> </em><em>remainder</em><em> </em><em>when</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>2</em><em>,</em><em> </em><em>checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>even</em><em> </em>

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>odd</em><em> </em><em>and</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>even</em><em> </em>

return gcd(a, b/2)

else :

return gcd(a, b-a)

<em>#since</em><em> </em><em>it's</em><em> </em><em>a</em><em> </em><em>recursive</em><em> </em><em>function</em><em>,</em><em> </em><em>it</em><em> </em><em>recalls</em><em> </em><em>the function</em><em> </em><em>with </em><em>new</em><em> </em><em>parameters</em><em> </em><em>until</em><em> </em><em>a</em><em> </em><em>certain</em><em> </em><em>condition</em><em> </em><em>is</em><em> </em><em>satisfied</em><em> </em>

print(gcd(124, 244))

print()

<em>#leaves</em><em> </em><em>a</em><em> </em><em>space</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>first</em><em> </em><em>output</em><em> </em>

print(gcd(4424, 2111))

Learn more :brainly.com/question/25506437

6 0
3 years ago
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