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3 years ago

The degree measure of two consecutive angles of a rhombus are represented by 5x+10 and 3x+50 find x

Mathematics

1 answer:

Readme [11.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

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KLMN is a parallelogram,

Aleonysh [2.5K]

Answer:

Ok soo.

Step-by-step explanation:

sum of two adjacent angles of parallelogram is 180

so angle K + angle L = 180

as AK bisects angle K and LA bisects angle L

so, angle AKL + angle ALK = 180 /2 = 90

in triangle AKL,

angle AKL + angle ALK+angle KAL = 180(sum of all angles in triangle = 180)

so, from above two equations,

angle KAL = 90

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3 years ago

What is <br> (3xy-1) (4xy+2)

zzz [600]

12x^2y^2+2xy-2 with problems like these use the app “Photomath” it will help you a lot.

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3 years ago

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A motorcycle is traveling at a constant speed of 75miles per hour. How many feet does it travel in 7seconds? Remember that 1 mil

Stella [2.4K]

770 feets per 7 seconds

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3 years ago

Write the expression <br> 3x-(2-4x) in standard form

Llana [10]

7x-2 is how it’s written in standard form

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3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

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

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago