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OleMash [197]
3 years ago
6

Lliana was part of a group that was working on changing 0.4 to fraction. Each member o the group had a different answer. Which a

nswer is correct?
A) 2/5
B) 4/10
Mathematics
2 answers:
Delicious77 [7]3 years ago
6 0

Answer:

Both are correct.

A) A decimal fraction

B) A common fraction

Step-by-step explanation:

<u>1 step.</u> Write down 0.4 divided by 1:

\dfrac{0.4}{1}.

<u>2 step.</u> Multiply both top and bottom by 10 (because there are 1 digit after the decimal point):

\dfrac{0.4}{1}=\dfrac{0.4\cdot 10}{1\cdot 10}=\dfrac{4}{10}.

This fraction is called a decimal fraction.

<u>3 step.</u> Simplify the decimal fraction:

\dfrac{4}{10}=\dfrac{2\cdot 2}{2\cdot 5}=\dfrac{2}{5}.

This fraction is called a common fraction.

inna [77]3 years ago
3 0

Answer:

The answer is C) 4/9

Step-by-step explanation:

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Marysya12 [62]

Answer:

hope it helps mark me brainlieast!

Step-by-step explanation:

<em>For triangle ABC with sides  a,b,c  labeled in the usual way, </em>

<em> </em>

<em>c2=a2+b2−2abcosC  </em>

<em> </em>

<em>We can easily solve for angle  C . </em>

<em> </em>

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<em> </em>

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<em> </em>

<em>C=arccosa2+b2−c22ab  </em>

<em> </em>

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<em> </em>

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<em> </em>

<em>−1≤cosC≤1  </em>

<em> </em>

<em>0∘≤C≤180∘  </em>

<em> </em>

<em>We needed to include the degenerate triangle angles,  0∘  and  180∘,  among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points. </em>

<em> </em>

<em>The Law of Cosines, together with  sin2θ+cos2θ=1 , is all we need to derive most of trigonometry.  C=90∘  gives the Pythagorean Theorem;  C=0  and  C=180∘  give the foundational but often unnamed Segment Addition Theorem, and the Law of Sines is in there as well, which I’ll leave for you to find, just a few steps from  cosC=  … above. (Hint: the Law of Cosines applies to all three angles in a triangle.) </em>

<em> </em>

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<em> </em>

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<em> </em>

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<em> </em>

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<em> </em>

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<em> </em>

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