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3 years ago

Angle A is a right angle in △ABC. Which side is opposite angle C? side BC side AB side AC

Mathematics

2 answers:

Hunter-Best [27]3 years ago

8 0

answer is side AB

I'm not 100% sure, just a guess

I hope it helps tho

vazorg [7]3 years ago

4 0

Answer:

Option 2nd is correct that is side AB.

Step-by-step explanation:

We are given a right angled ΔABC in which ∠A is right angle.

We have to find which side is opposite to ∠C

Since, ∠A is right angle it implies side CB is hypotenuse in ΔABC.

We know that side which are opposite to a angle does not includes the vertex of that angle.

like above hypotenuse is opposit of ∠A and name of hypotenuse does not have vertex a in it.

Similarly Side opposite to ∠C will not have vertex C in it.

So, AB is the side opposite to ∠C.

Therefore, Option 2nd is correct that is side AB.

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3 years ago

Will mark brainiest for simplest and most coherent answer. What does it mean if an x (input) or y (output) value doesn't make se

Maksim231197 [3]

Answer:

I tried to answer but I still would like to know what doesn't make sense.

Considering the identity:

$\sec^2\theta-\tan^2\theta=1 \implies |\sec\theta|\geq1$

$\forall\;\theta\in\mathbb{R}-\{(2n+1)\frac{\pi}{2}, n\in\mathbb{Z} \}$

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8 0

3 years ago

Select all the correct answers.<br> In which pairs of matrices does AB = BA?

horsena [70]

In order to multiply a matrix by another matrix, we multiply the rows in the first matrix by the columns in the other matrix (How this is done is shown below)

To determine the pairs of matrices that AB=BA, we will determine AB and BA for each of the options below.

For the first option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-2\times5)+(1\times3)&(-2\times0)+(1\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-10+3&0+2&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&-7&2&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-2)&(5\times0)+(0\times 1)&(3\times1)+(2\times-2)&(3\times0)+(1\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-4&0+2&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&-1&2&\end{array}\right] \\$

∴ AB≠BA

For the second option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-1\times3)+(2\times6)&(-1\times0)+(2\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-3+12&0+-6&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-1)&(3\times0)+(0\times 2)&(6\times1)+(-3\times-1)&(6\times0)+(-3\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+3&0+-6&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

Here AB = BA

For the third option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-1\times5)+(2\times3)&(-1\times0)+(2\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-5+6&0+4&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-1)&(5\times0)+(0\times 2)&(3\times1)+(2\times-1)&(3\times0)+(2\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-2&0+4&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

Here also, AB=BA

For the fourth option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-2\times3)+(1\times6)&(-2\times0)+(1\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-6+6&0+-3&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&0&-3&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-2)&(3\times0)+(0\times 1)&(6\times1)+(-3\times-2)&(6\times0)+(-3\times1)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+6&0+-3&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&12&-3&\end{array}\right] \\$

Here, AB≠BA

Hence, it is only in the second and third options that AB = BA

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right] B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$ and $A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

Learn more on matrices multiplication here: brainly.com/question/12755004

8 0

3 years ago

Read 2 more answers

Pretty please choose a correct answer, thank youuuu

klasskru [66]

The correct answer is b

7 0

4 years ago

Read 2 more answers