How to find quartiles of a matrix column in r?

Help<br><br> Determine which equation is parallel to line jk and which is perpendicular to line jk.

alukav5142 [94]

Answer:

5x + 3y = 13 parallel.

6x - 10y = 7 perpendicular.

Step-by-step explanation:

Two lines with slopes $m_1$ and $m_2$ are parallel when $m_1=m_2$ and are perpendicular when $m_1*m_2=-1$

Now determine the slope of all the lines

The line jk passes through the points

(-5,5) and (1,-5) so its slope is

$\frac{-5-5}{1-(-5)}=\frac{-10}{6}=-\frac{5}{3}$

To determine the slope of the lines in the blue rectangles, isolate y from each one and the coefficient of x is the slope

5x-3y = 8 ------> y = (5/3)x + 8/3 ------> slope 5/3

Neither parallel nor perpendicular.

6x+10y = 11 ------> y = (-6/10)x + 11/10 = (-3/5)x + 11/10 ------> slope -3/5

Neither parallel nor perpendicular.

5x + 3y = 13 ------> y = (-5/3)x + 13/3 ------> slope -5/3

This line is parallel

6x - 10y = 7 ------> y = (6/10)x - 7/10 = (3/5)x -7/10 ------> slope 3/5

Since (-5/3)(3/5) = -1 this line is perpendicular.

8 0

3 years ago

Two fractions one positive one negative whose product is 5/8

Ray Of Light [21]

-10/8 and + 15/8. Those should work.

3 0

3 years ago

Please help with this problem

Dennis_Churaev [7]

Answer:

that looks hard

Step-by-step explanation:

that's what she said

lol

4 0

3 years ago

Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clu

dmitriy555 [2]

Answer:

The figure is attached and the total distance is 1031 feet.

Step-by-step explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

$d_{T}=[d_{FP\ to\ Clue1}+d_{Clue1\ to\ Clue2}+d_{Clue2\ to\ Clue3}]*distance\ per\ unit\\d_{T}=[\sqrt{(Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2}+\sqrt{(Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2}+\sqrt{(Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2}]**distance\ per\ unit\\$ Substituting the values

$d_{T}=[\sqrt{(0-0)^2+(5-0)^2}+\sqrt{(6-0)^2+(0-5)^2}+\sqrt{(0-6)^2+(-5-0)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{(0)^2+(5)^2}+\sqrt{(6)^2+(-5)^2}+\sqrt{(-6)^2+(-5)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{0+25}+\sqrt{36+25}+\sqrt{36+25}]*50 \text{ feet}\\d_{T}=[\sqrt{25}+\sqrt{61}+\sqrt{61}]*50 \text{ feet}\\d_{T}=[5+7.81+7.81]*50 \text{ feet}\\d_{T}=[20.62]*50 \text{ feet}\\d_{T}=1031 \text{ feet}\\$

So the total distance travelled is 1031 feet.

4 0

4 years ago

Assume ∆ EFG ≅ ∆ JKL. If m∠J =90, m∠K =26, what is the measure of ∠E ?

Anna11 [10]

Answer:

90 A pex

Step-by-step explanation:

8 0

3 years ago