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3 years ago

Two angles of a triangle measures of 70° and 85° which is not the measure of an exterior angle of the triangle

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

7 0

I hope this helps you

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Distributing property and combining like terms <br> -10(1-3x)-10

ikadub [295]

Answer:

-20 + 30x

Step-by-step explanation:

-10(1) = -10

-10(-3x) = 30x

new equation

-10 + 30x -10

-10 + -10 = -20

-20 + 30x

7 0

3 years ago

How to solve the equation and rounding the answer to four decimal places

pogonyaev

The answer is 0.5778 or 0.578 rounded

5 0

3 years ago

Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)

Yakvenalex [24]

Answer:

$\sin d = \frac{4}{7}$ ; $\sin e = \frac{\sqrt{33} }{7}$

$\cos d = \frac{\sqrt{33} }{7}$ ; $\cos e = \frac{4}{7}$

$\tan d = \frac{4}{\sqrt{33} }$ ; $\tan e = \frac{\sqrt{33} }{4}$

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

$\sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\cos \alpha = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\tan \alpha = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }$

__________________________________________________

According to the question ,

1) When α (alpha) = d

$\sin d = \frac{4}{7}$

$\cos d = \frac{\sqrt{33} }{7}$

$\tan d = \frac{4}{\sqrt{33} }$

2) When α (alpha) = e

$\sin e = \frac{\sqrt{33} }{7}$

$\cos e = \frac{4}{7}$

$\tan e = \frac{\sqrt{33} }{4}$

3 0

3 years ago

-3(-7) answer to this please

Naddika [18.5K]

Answer:

Step-by-step explanation:

Negative times negative is positive.

-3(-7) = 21

8 0

3 years ago

Please help if you can pic attatched

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Subtract row 3 from row 1:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]$

Subtract row 3 from 7 times row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]$

Divide row 3 by 17:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 2 of row 3 from row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 6 of row 2 from row 1:

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

$C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]$

$C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]$

3 0

3 years ago