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3 years ago

The endpoints of EFare E(xE , yE) and F(xF , yF). What are the coordinates of the midpoint of EF?

Mathematics

2 answers:

jekas [21]3 years ago

3 0

<span>The coordinates of the midpoint of EF is ((xE + xF)/2, (yE + yF)/2)
The third option is the correct answer.
</span>

algol133 years ago

3 0

Answer:

(C)

Step-by-step explanation:

It is given that the end points of EF are $E(x_{E},y_{E})$ and $F(x_{F},y_{F})$ .

Let the mid point of EF be M, then the coordinates of M can be found by using the mid point formula,

M= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

M= $(\frac{x_{E}+x_{F}}{2},\frac{y_{E}+y_{F}}{2})$

Thus, option C is correct.

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Lupita rides a taxi that charges a flat rate of $6.75 plus $3.20 per mile. If the taxi charges Lupita $40.03 in total for her tr

Alex777 [14]

Hello! In order to understand this question, we need to take a look at the content that is involved.

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8 0

3 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0

1 year ago

32%

lisabon 2012 [21]

At least 4

10% of 12 is 1.2

So 30% is 3.6

And 1% 0.12

So 2% is 0.24

Plus 3.6 onto 0.24 and you get 3.84

4 0

3 years ago