In ΔHIJ, the measure of ∠J=90°, the measure of ∠I=52°, and JH = 4.2 feet. Find the length of IJ to the nearest tenth of a foot.

Mathematics

1 answer:

balandron [24]3 years ago

5 0

Answer:3.3

Step-by-step explanation:

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Find missing length

svetlana [45]

By the Pythagorean theorem
.. 15^2 = 10^2 +a^2
.. 225 = 100 +a^2
.. 125 = a^2
.. √125 = a
.. 5√5 = a ≈ 11.18
The missing side length is 5√5 cm.

6 0

3 years ago

2 If A= 4 3 find A1 using elementary row operations.

MA_775_DIABLO [31]

Answer: $A^{-1}=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$

Step-by-step explanation:

$\left[\begin{array}{cc}2&1\\4&3\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$\dfrac{1}{2}Row\ 1\rightarrow\left[\begin{array}{cc}1&\frac{1}{2}\\4&3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\0&1\end{array}\right]$

$Row\ 2 -4 \ Row\ 1\rightarrow \left[\begin{array}{cc}1&\frac{1}{2}\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&0\\-2&1\end{array}\right]$

$Row\ 1-\dfrac{1}{2}\ Row\ 2 \rightarrow \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}&-\frac{1}{2}\\-2&1\end{array}\right]$