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3 years ago

F(x)=x²+2x-5 determine the vertex

Mathematics

2 answers:

Sliva [168]3 years ago

5 0

Answer:

the point -5 is the vertex on the graph because it is the y-intercept.

Step-by-step explanation:

Finger [1]3 years ago

3 0

Answer:-16

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Evaluate this expression |-9|

N76 [4]

Answer:9

Step-by-step explanation:

The distance between-9 and 0 is 9.

3 0

3 years ago

Read 2 more answers

If y varies directly as x and y=32 when x=4 find the value of y when x=5

Alex Ar [27]

Answer:

y=40

Step-by-step explanation:

32 divided by 4 equals 8. And 5 times 8 equals 40.

3 0

3 years ago

PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION

Alex17521 [72]

Solution:

$\sqrt{4x + 13} = x + 2$

First square both sides.

$= > ( \sqrt{4x + 13} ) ^{2} = (x + 2) ^{2}$

Now, square root and square gets cancel out in the LHS. And in the RHS, apply the identity: (a + b)² = a² + 2ab + b².

$= > 4x + 13 = {(x)}^{2} + 2 \times x \times 2 + (2) ^{2} \\ = > 4x + 13 = {x}^{2} + 4x + 4$

Now, transpose 4x and 4 to LHS.

$= > 4x - 4x + 13 - 4 = {x}^{2} \\$

Now, do the addition and subtraction.

$= > {x}^{2} = 9 \\ = > x = \sqrt{9} \\ = > x = ±3$

Answer:

x = ± 3

Hope you could understand.

If you have any query, feel free to ask.

3 0

2 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

What is two the two this is Free poin ]t because I'm nice

satela [25.4K]

Two the two is the four.

4 0

2 years ago