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BigorU [14]
4 years ago
13

Is the following relation a function???

Mathematics
1 answer:
bulgar [2K]4 years ago
7 0
No i do not believe so
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Miranda says that 30x26 is greater than 20x36 is she correct ?
julsineya [31]
Miranda is correct. 30×26=780 and 20×36=720 so 30×26>20×36
4 0
3 years ago
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I need help with this problem, if anyone could help ASAP, that would be much appreciated. In the figure below, mROP = 125° Find
VARVARA [1.3K]

Answer:

mRP = 125°

mQS = 125°

mPQR = 235°

mRPQ = 305°

Step-by-step explanation:

Given that

  • mROP = 125°
  • ∠ROP is a central angle

Then:

  • measure of arc RP, mRP = mROP = 125°

Given that

  • ∠QOS and ∠ROP are vertical angles

Then:

  • mQOS = mROP = 125°
  • measure of arc QS, mQS = mROP = 125°

Given that

  • ∠QOR and ∠SOP are vertical angles

Then:

  • mQOR = mSOP

Given that

  • The addition of all central angles of a circle is 360°

Then:

mQOS + mROP + mQOR + mSOP = 360°

250° + 2mQOR = 360°

mQOR = (360°- 250°)/2

mQOR = mSOP = 55°

And (QOR and SOP are central angles):

  • measure of arc QR, mQR = mQOR = 55°
  • measure of arc SP, mSP = mSOP = 55°

Finally:

measure of arc PQR, mPQR = mQOR + mSOP + mQOS = 55° + 55° + 125° = 235°

measure of arc RPQ, mRPQ = mROP + mSOP + mQOS = 125° + 55° + 125° = 305°

6 0
3 years ago
Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why th
Veseljchak [2.6K]

If the coefficient matrix has a pivot in each column, it means that it is shaped like this:

A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]

So, the correspondant system

Ax = b

will look like this:

\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]

This turn into the following system of equations:

\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}

The last equation is solvable for x_4: we easily have

x_4=\dfrac{b_4}{a_{4,4}}

Once the value for x_4 is known, we can solve the third equation for x_3:

x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}

(recall that x_4 is now known)

The pattern should be clear: you can use the last equation to solve for x_4. Once it is known, the third equation involves the only variable x_3. Once

4 0
4 years ago
PLEASE PLEASE HELP MEH
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Answer is in bold and i will add step
Since they are vertical angles they will be the same. Since 6=146 2=146

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