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Usimov [2.4K]
3 years ago
10

two angles are complementary they also have the same measurements which statements below are correct ​

Mathematics
1 answer:
podryga [215]3 years ago
3 0
The first both angles measure 45 degrees




Step by step explanation:


Complementary angles are angles that add up to 90 degrees and it said they both are equal which means both of them have to be half of the 90 degrees and half of 90 is 45 so 45 degrees
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Can a common denominator of mixed numbers be 20?
gladu [14]
Yes, a common denominator is just the smallest multiple of all numbers in a group. 

So the common denominator of 4, 8 and 12 would be 4, because 4 goes into all those numbers.
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18. Paul can type 60 words per minute and Jennifer can type 80 words per minute. How does Paul's typing speed compare to Jennife
Serggg [28]
Paul can type fast but 20 words less than Jennifer a minute.
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1/2+1/9Please help me
musickatia [10]

<em>If the fraction whose denominator are equal then they will add up</em>

In the given fraction 1/2 +1/9, the denominator of both the fraction 1/2 & 1/9 is not same

so, to make the base same we take the LCM of the 2 & 9

\begin{gathered} \text{LCM of 2 \& 9 is 18} \\ Si,\text{ the fraction will be :} \\ \frac{1}{2}+\frac{1}{9}=\frac{9+2}{18} \\ \frac{1}{2}+\frac{1}{9}=\frac{11}{18} \end{gathered}

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3 0
1 year ago
Find the altitude of the triangle with a side of 18.<br> Put into sqrt.
ira [324]
<h2>15 58</h2>

Step-by-step explanation:

BD = DC = 1/2 BC ---> AD is median ( because in equilateral triangle altitude equals to median )

so, therefore

AB = 18 (given)

BD = 1/2 BC

BD = 1/2 × 18

BD = 9

AB² = AD² + BD² ---( by pythagoras theorm)

18² = AD² + 9²

324 = AD² + 81

324 - 81 = AD²

243 = AD²

then,

AD = √243

AD = 15 58

<h2>MARK ME BRAINLIST</h2><h3>PLZ FOLLOW ME</h3>

3 0
3 years ago
If A is a square matrix such that A^2=2I,then A^-1=?​
Afina-wow [57]

A^2=2I

Recall that I=AA^{-1}=A^{-1}A, so

A^2=2(A^{-1}A)

Multiply on the right by A^{-1} and regroup terms:

A^2A^{-1}=2(A^{-1}A)A^{-1}

A(AA^{-1})=2A^{-1}(AA^{-1})

A=2A^{-1}

and finally solve for A^{-1}:

\boxed{A^{-1}=\dfrac12A}

3 0
3 years ago
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