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Ksivusya [100]
4 years ago
6

What do all 3 forms of the quadratic equation (General/standard, vertex, factor) have in common?

Mathematics
2 answers:
Tasya [4]4 years ago
8 0

Answer:


Step-by-step explanation:

The equations have the same slope and same y-intercept; therefore, they are the same line.

finlep [7]4 years ago
7 0

All three forms of the equation have the same two solutions.

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15 POINTS PLEASE HELP!!! WILL MARK BRAINLIEST
dangina [55]

Given: ∠ DEF

To construct: ∠TSZ ≅ ∠DEF

Construction: Consider the attachment

Step-01: Draw a line XY and choose a point S on it as a vertex of the required angle. Further marks point T such that DE = ST

Step-02: Take an arc AB from point E in ∠DEF of any length and draw at point S which cuts at point P on XY line.

Step-03: Take another arc of length AB from point B in ∠DEF and draw from point P which cuts to the previous arc at Q.

Step-04: Now, join the point SQ and extend up to Z such that EF = SZ

Hence, ∠ TSZ will be the required congruent constructed angle to∠DEF

6 0
3 years ago
Please help me!! Just 13-18
Nady [450]

Answer:

-5

Step-by-step explanation:


6 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0
3 years ago
Please help meeeeeee​
amm1812

Answer:

y=2/3x +5

Step-by-step explanation:

to find the gradient

9--1/6--9

10/15

2/3

y-y1=m(x-x1)

y-9=2/3(x-6)

y-9=2/3x-4

y=2/3x+9-4

y=2/3x+5

4 0
4 years ago
Set up the appropriate function to be maximized or minimized.
rosijanka [135]
I would have to say is $14 per foot
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3 years ago
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