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joja [24]
3 years ago
11

Find the length of the other two side of the isosceles right triangle

Mathematics
1 answer:
Valentin [98]3 years ago
4 0

Answer:

The other two sides are 7 units each.

Step-by-step explanation:

The triangle is isosceles.

The sides are in the ratio of

x : x : x \sqrt{2}

The hypotenuse is

7\sqrt{2}=x \sqrt{2}\\\\x = 7

So, the other two sides are 7 units each.

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(1,3), m = - 3/4 using that as a fraction write an equation in point-slope form
sukhopar [10]
Y=mx+b   m=-3/4  (1,3)
3=-3/4(1)+b
3=-3/4+b
b=3 3/4

y=-3/4+3 3 3/4
7 0
3 years ago
Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1
sasho [114]
I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse. 

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.) 

\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1

Where (x_{1},y_{1}) is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is: 

(-3,5)

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis. 

The major axis in this case is that of the y-axis. In other words, 

b^2>a^2 

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are: 

(-3,10)
 & (-3,0)

I really hope this helped you! (Partially because I spent a lot of time on this lol) 

Sincerely,

~Cam943, Junior Moderator
6 0
3 years ago
I will mark brain Can someone help me do not answer if u dont know
meriva

Answer:

1. y = 1/3x+-1

2. y = 4

Hope this helps man :D

6 0
3 years ago
Read 2 more answers
CAN SOMEONE HELPPPPPP PLEASE?!??!!!!?
iris [78.8K]
If AM=10, then AC=10 because both are radii. TC is a diameter so that’ll just be the radius doubled. TC=20!
8 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Match the systems of equations with their solutions.
Oksanka [162]

Answer:

See explanation for matching pairs

Step-by-step explanation:

Equations

(1)

x - y = 25

2x + 3y = 180  

(2)

2x - 3y = -5

11x + y = 550  

(3)

x - y = 19

-12x + y = 168

Solutions

(-17,-36)

(47, 33)

(51, 26)

Required

Match equations with solutions

(1) x - y = 25 and 2x + 3y = 180

Make x the subject in: x - y = 25

x = 25 + y

Substitute x = 25 + y in 2x + 3y = 180

2(25 + y) + 3y = 180

50 + 2y + 3y = 180

50 + 5y = 180

Collect like terms

5y = 180-50

5y = 130

Solve for y

y =26

Recall that: x = 25 + y

x = 25 + 26

x = 51

So:

(x,y) = (51,26)

(2)  2x - 3y = -5 and 11x + y = 550

Make y the subject in 11x + y = 550

y = 550 - 11x

Substitute y = 550 - 11x in 2x - 3y = -5

2x - 3(550 - 11x) = -5

2x - 1650 + 33x = -5

Collect like terms

2x  + 33x = -5+1650

35x = 1645

Solve for x

x = 47

Solve for y in y = 550 - 11x

y = 550 - 11 * 47

y = 550 - 517

y = 33

So:

(x,y) = (47,33)

(3)

x - y = 19  and -12x + y = 168

Make y the subject in -12x + y = 168

y = 168 + 12x

Substitute y = 168 + 12x in x - y = 19

x - 168 - 12x = 19

Collect like terms

x -12x = 168 + 19

-11x = 187

Solve for x

x = -17

Solve for y in y = 168 + 12x

y =168-12 *17

y =-36

So:

(x,y) = (-17,-36)

4 0
2 years ago
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