All categories

3 years ago

Can somebody please help me with this question!

Mathematics

2 answers:

uranmaximum [27]3 years ago

7 0

Step-by-step explanation:

angle A = angle D

angle C common angle

angle E = 180 - 63 - 52

5x = 65

x = 13deg

Topic: congruence and similarity

If you like to venture further, do check out my insta (learntionary) where I regularly post useful math tips! Thank you!

Lelu [443]3 years ago

7 0

Answer:

x = 13°

Step-by-step explanation:

Because of the given congruency,

Angle A = Angle D = 63°.

Angle C = 52°

Angle B = Angle E = 5x

Now, In Triangle ABC,

Angle A + Angle B + Angle C = 180°.

63° + 5x + 52° = 180°

115° + 5x = 180

5x = 65°

x = 13°

HOPE IT HELPS!!!

You might be interested in

PLEASSSSSEEE HELPPP ME

Sonja [21]

I hope this helps you

3 0

3 years ago

Which equation correctly applies the distributive property?

yuradex [85]

<span> −3⋅(6.48)=(−3⋅6)+(−3⋅0.4)+(−3⋅0.08) </span>correctly applies the distributive property

8 0

4 years ago

Read 2 more answers

What is 3 to the power of 3 over 2 equal to?

ddd [48]

27/2. 3x3x3 is 27. .

5 0

3 years ago

Read 2 more answers

PLS help with 2,4,6 I rlly need help understanding help guide me through

valentina_108 [34]

Try to understand this. Hope this help. Answer for 2 only.

3 0

3 years ago

Read 2 more answers

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

6 0

4 years ago