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andreyandreev [35.5K]

3 years ago

8

Best explained and correct answer gets brainliest.

2 answers:

babymother [125]3 years ago

7 0

The answer is B ... proven by AAS

leva [86]3 years ago

5 0

Hey!

____________________________

Given-:

∆RQT
QS bisects ∠RQT
∠R= ∠T

To prove- RS=TS

Proof-

∠R= ∠T [Given]
∠RQS= ∠TQS [ QS bisects ∠RQT]
RQ= QT [∠R= ∠T]

Here we get, ∆RSQ ≅ ∆TSQ

Thus, By C.P.C.T, we get,

RS=TS

Thus, correct Option=> (B)
____________________________

Hope it helps...!!!

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7 0

3 years ago

A rectangle has a height of 3 and a width of 2x*2 + 3x - 5 , express area of whole rectangle, please answer this ASAP !!!

monitta

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8 0

4 years ago

For a certain horse race comma the odds in favor of a certain horse finishing in second placehorse race, the odds in favor of a

NikAS [45]

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3 years ago

If A is nonsingular, then (AT )-1 =(A-1) T . (a) Verify this theorem for 2 × 2 matrices

kipiarov [429]

Answer:

Verified

Step-by-step explanation:

Let the 2x2 matrix A be in the form of:

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

Where det(A) = ad - bc # 0 so A is nonsingular:

Then the transposed version of A is

$A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]$

Then the inverted version of transposed A is

$(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

The inverted version of A is:

$A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]$

The transposed version of inverted A is:

$(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

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$(A^T)^{-1} = (A^{-1})^T$

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3 0

3 years ago