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3 years ago

Fidn the points of intersection

Mathematics

1 answer:

Sedbober [7]3 years ago

8 0

EQ. 1: x + y = 8

EQ. 2: 4x-y = 7

Rewrite EQ. 1 as EQ. 3: x = 8-y

Replace x in EQ.2 with EQ. 3:

4(8-y) - y = 7

Use the distributive property:

32 - 4y - y = 7

Combine like terms:

32 - 5y = 7

Subtract 32 from each side:

-5y = -25

Divide both sides by -5

y = -25 / -5

y = 5

Now replace y with 5 in EQ. 3 to solve for X:

x = 8-5

x = 3

The point of intersection is X = 3, Y = 5, which is written as (3,5)

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Dinesh has been keeping track of his time at work over the past month. Answer the questions below using the information provided

Gala2k [10]

The total hours that Dinesh worked each week over the past month are as follows:

<h3>Dinesh's Time in Hours</h3>

Total Hours

Worked

Week 1 37.500

Week 2 36.071

Week 3 35.999

Week 4 34.570

Total 144.140

<h3>Data and Calculations:</h3><h3>Dinesh's Time in Hours</h3>

Meetings Administration Project X - Project X- Total Hours

Design Deliverables Worked

Week 1 7¹/₄ 4¹/₄ 23¹/₃ 2²/₃

Week 2 4²/₃ 2⁴/₇ 16¹/₃ 12¹/₂

Week 3 3.50 3.83 9.33 19.33

Week 4 6.17 3.40 4.20 20.80

<h3>Conversion into Dinesh's Time in Hours (decimals)</h3>

Meetings Administration Project X - Project X- Total Hours

Design Deliverables Worked

Week 1 7.250 4.250 23.333 2.667 37.500

Week 2 4.667 2.571 16.333 12.500 36.071

Week 3 3.500 3.833 9.333 19.333 35.999

Week 4 6.170 3.400 4.200 20.800 34.570

Total 21.587 14.054 53.199 55.300 144.140

Thus, Dinesh's total hours worked for the past month was <u>144.140 hours</u>.

Learn more about calculating hours worked at brainly.com/question/17054097

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8 0

2 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

Factorise :x^2y^2-16

Lyrx [107]

Answer:

(xy - 4)(xy + 4).

Step-by-step explanation:

The is the difference of 2 squares.

a^2 - b^2 = (a - b)(a + b: so we have:

x^2y^2 - 16 = (xy - 4)(xy + 4)

5 0

3 years ago

Is .32903 greater than 0.59

leonid [27]

No, <span>.32903 is not greater than 0.59</span>

5 0

3 years ago

Read 2 more answers

Find the output, h, when the input, x, is - 18.<br><br> h=17+ x/6

Nookie1986 [14]

Answer:

h = 20

Step-by-step explanation:

h = 17 + x/6

h = 17 + (18) / 6

h = 17 + 3

h = 20

5 0

2 years ago

Read 2 more answers