Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AD = 4

Mathematics

1 answer:

Vinil7 [7]2 years ago

4 0

Answer:

BD = 6

Step-by-step explanation:

Theorem: The altitude to the hypotenuse of a right triangle is the mean proportional between the segments of the hypotenuse.

So, $\frac{4}{BD} = \frac{BD}{9}$

$BD^{2} = 36$

BD = 6

See how easy that is IF you know the theorem.

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Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Thepotemich [5.8K]

$\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

Explanation

$\frac{1}{a+bi}=c+di$

Step 1

multiplicate by the conjugate

$\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}$

notice that

$\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\ \end{gathered}$ $\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

I hope this helsp you

6 0

1 year ago

What is 1m + 3/4 this too hard

TiliK225 [7]

Answer:

m = - 3 / 4

Step-by-step explanation:

1m + 3/4 = 0

m = - 3 / 4

hope this helps

4 0