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12345 [234]
3 years ago
13

Steve built a square hamster pen that has a perimeter of 240 centimeters.

Mathematics
1 answer:
vesna_86 [32]3 years ago
6 0

Answer:

What's the question?

Step-by-step explanation:

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Step-by-step explanation:


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3 years ago
In triangle △JKL, ∠JKL is right angle, and KM is an altitude. JL=25 and JM=5, find KM.
d1i1m1o1n [39]

Answer:

KM = 15

Step-by-step explanation:

From the diagram, ΔJKL and ΔJKM are similar to each other because they share the same angle J and they are both right angle triangle. Therefore they are similar by AA property.

Since JL=25 and JM=5, JM = JL + JM= 25 + 5 = 30

Since ΔJKL and ΔJKM are similar to each other, therefore:

\frac{JK}{JL}=\frac{JM}{JK}\\  Substituting:\\\frac{JK}{30}=\frac{5}{JK}\\JK^2=150\\JK=\sqrt{150}\\ \\From\ hypotenuse:\\JK^2=JM^2+KM^2\\Substituting:\\(\sqrt{150} )^2=5^2+KM^2\\150=25+KM^2\\KM^2=150-25=125\\KM=\sqrt{125}\\ KM=15

4 0
4 years ago
Perform the indicated operation and express the result as a simplified complex number. I need help with 35
skelet666 [1.2K]

Given:-

\frac{3+4i}{2-i}

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

2+i

So now we multiply. we get,

\frac{3+4i}{2-i}\times\frac{2+i}{2+i}

Now we simplify. so we get,

\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}

We know the value of,

i^2=-1

Substituting the value -1. we get,

\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{                               =}\frac{6-4+11i}{2+1} \\ \text{                              =}\frac{2+11i}{3} \end{gathered}

So now we split the term to bring it into the form a+ib. so we get,

\frac{2+11i}{3}=\frac{2}{3}+i\frac{11}{3}

So the required solution is,

\frac{2}{3}+i\frac{11}{3}

3 0
1 year ago
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