Ask question

Ask question

All categories

2 years ago

7

How do i solve for angle 1 and 2?

2 answers:

Drupady [299]2 years ago

6 0

Both angles would be 118° hope this helped and I hope you consider giving me brainliest

nekit [7.7K]2 years ago

5 0

If you are reading this have a good day you might be down but you will make it your still young remember that

You might be interested in

See attachment. What would be the perpendicular lines of this sense there are 4 right anglrs being formed? I thought perpendicul

chubhunter [2.5K]

Yeah not just one letter but two letters perpendicular to another two letters. Each 2 letters like AB is a straight line which is crossed by another 2 letters making it perpendicular.

4 0

2 years ago

I really NEED HELP<br> Ill give you a brainiest if you guys help me ASAP because I dont understand

MAXImum [283]

Answer:

<h2>I think it is B if I am wrong i am <u><em>sooooo</em></u> sorry!</h2>

Step-by-step explanation:

<h2 />

8 0

1 year ago

Read 2 more answers

The balance in a student’s bank account is $1000. Each month, $150 is withdrawn.

UNO [17]

Answer:

I think its 900

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Gordan's front lawn measures 2,655 square feet. It takes him 9 minutes to mow it on his riding lawn mower. How many square feet

zlopas [31]

It would be 295 square feet.

5 0

2 years ago

Read 2 more answers

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

2 years ago