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Mashcka [7]
3 years ago
12

Find the value of 5.8 ÷ 1.2.

Mathematics
1 answer:
EastWind [94]3 years ago
5 0
Your answer is 4.83
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Which pairs of angles in the figure below are vertical angles?<br><br> Check all that apply.
erica [24]
I think it’s all of them.
7 0
4 years ago
Read 2 more answers
Help please!!
yanalaym [24]

Answer:  (1,-1)

Step-by-step explanation:

Midpoint of BC=(6+4)/2, (3–1)/2. =(5,1)

Slope of BC is (3+1)/4–6)= 4/-2 = -2

Slope of perpendicular bisector of BC =+1/2

Eqn of perpendicular bisector is : Y-1 =1/2 (x-5)

Y=1/2 •(x-5) +1

Midpoint of AB. (6–2)/2, (3–1)/2 ={2,1)

Slope of AB is(3+1)/(-2–6) = 4/-8 =-1/2

Slope of perpendicular bisector = +2

Eqn of perpendicular bisector is Y-1. =2( X-2)

Y=2X-4+1 = 2X -3

Solving Y=(X-5)/2 +1

& Y=2X-3

2X-3 =(x-5/2)+1

2X-4 =(x-5/2)

4X-8 = x-5

3X =3

X=1

Y= 2×1–3= -1

Circumcentre is(1,-1)

8 0
3 years ago
What is the least common multiple you could use to find 4/5 + 5/6?
damaskus [11]
Tbh idk ask your teacher
4 0
4 years ago
Need to know this now pls !!! NEED THE PROCESS , TOO!!!<br><br> x squared - 18x = -117
nydimaria [60]

Answer:

x = 9 + 6i or x = 9 - 6i

Step-by-step explanation:

x² - 18x + 117 = 0

Complete the square and solve:

(x - 9)² - 81 + 117 = 0

(x - 9)² + 36 = 0

(x - 9)² = -36

x - 9 = ±sqrt(-36)

x - 9 = ±sqrt(-1).sqrt(36)

i = imaginary number = sqrt(-1)

x - 9 = ±6i

x = 9 ± 6i

There are no real solutions, only imaginary ones

9 + 6i and 9 - 6i

3 0
3 years ago
Read 2 more answers
Help me pleaseeeee.
Natali [406]

Answer:

7.) Line, point

8.) Line

9.) Line, point, rotational

10.) Rotational

11.) None

12.) Line

4 0
3 years ago
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