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Ghella [55]
3 years ago
15

Can you find the missing angles?

Mathematics
2 answers:
S_A_V [24]3 years ago
8 0

There are 180<span>°<span>(180 degrees) in a triangle.
</span>
We are given one angle, in two triangles(essentially one), but there are two "hidden angles" that are not listed.
 
A 90</span>° angle is formed when two lines/rays/etc. intersect perpendicular to each other. So now we know two angles in triangle yellow, and one angle in triangle purple. 
<span>
Since we know a triangle has 180 degrees, 180 - 90 - 45 = 45 degrees.

We have all three angles for triangle yellow.

Now, I don't have any proof unfortunately, but I believe the remaining angles for triangle purple is 30 degrees and 60 degrees.

TRIANGLE YELLOW:
45, 90, 90

TRIANGLE PURPLE(NO PROOF):
30, 60, 90</span>
Nitella [24]3 years ago
5 0
Every triangle has 180 degrees when you add there angles so you know that they will add up to 180
they are right triangles so the right angle has 90 degrees 
so for the yellow triangle 
90+45+x=180
135+x=180
45=x
we can tell where the purple and yellow triangle touch is 90 so 45+x=90 to find that angle which is 45
and the other one is 45 
so there are two 90 degree angles and three 45 degree angles.
I hope that made sence
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Answer:

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Step-by-step explanation:

Given the matrix

A=\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right]

Calculate A² - 6A + 11 I

A^2 = A*A= \left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] *\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] = \left[\begin{array}{ccc}2*2-3*1&2*3+3*4\\-1*2-4*1&-1*3+4*4\end{array}\right] =\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right]

6A=6*\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] =\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right]

11 I = 11 * \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =\left[\begin{array}{ccc}11&0\\0&11\end{array}\right]

∴ A² - 6A + 11 I = \left[\begin{array}{ccc}1&18\\-6&13\end{array}\right] -\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right] +\left[\begin{array}{ccc}11&0\\0&11\end{array}\right] =\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]

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Answer:

<h2><em><u>-9</u></em></h2>

Step-by-step explanation:

-  |3a +  \sqrt{b} |

<em><u>By</u></em><em><u> </u></em><em><u>putting</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>values</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>-4</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>b</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>9</u></em><em><u>,</u></em>

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= -(9)

= <em><u>-9 (Ans)</u></em>

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