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blagie [28]
2 years ago
15

HEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLP

Mathematics
2 answers:
Marina CMI [18]2 years ago
5 0

Answer:

your answer would be 6

Step-by-step explanation:

(3)4-6

12-6

6

Roman55 [17]2 years ago
3 0

Answer:

I think it would be 18

Step-by-step explanation:

3 x 4 =12

12 + 12= 24

24 -6=18

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When constructing a perpendicular bisector, why must the compass opening be greater the 1/2 than the length of the segment?
Alex787 [66]

Answer.

It is Important to open the compass more than half the length so that the arcs made would meet.

Explanation.

When you want to construct a perpendicular line to a line segment, say line AB, the follow steps are taken;

1) Open your compass to a radius greater than half the line AB.

2) With A as your center make an arc on top and below line AB.

3) Using the same radius and with B as our center, make other arcs on top and below line AB to meet the first arcs on both sides of the line.

4) Join the two intersection of the two arcs to cut line AB at M.

In so doing the line AB is said to be bisected perpendicularly. If the compass was opened less than half line AB, then the arcs could not have meet. 

8 0
3 years ago
HELPPP PLS
weqwewe [10]

Answer:

It would be (0,0)

Step-by-step explanation:

7-7=0 and 3.5-3.5=0

6 0
3 years ago
A line passes through the point (-8,8) and has a slope of -5/4
mariarad [96]
What are you looking for here?
4 0
3 years ago
Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0
gtnhenbr [62]

Answer:

Step-by-step explanation:

Given is a differential equation as

ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0

Divide this by t to get in linear form

y'+6y/t = t-1+1/t

This is of the form

y' +p(t) y = Q(t)

where p(t) = 1/te^(\int 1/tdt) = t

So solution would be

yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C

siubstitute y(1) = 16

16 = 16^3/3-128+1+C\\C = -1206

4 0
3 years ago
I need help with this ASAP plzzzzz
miss Akunina [59]

Answer:

23. \frac{4\sqrt{3} }{3}

24. \frac{7\sqrt{3}}{3}

25. 6\sqrt{3}

26. \frac{10\sqrt{3} }{3}

27.  14

28. 4\sqrt{3}

Step-by-step explanation:

To solve these i used SOHCAHTOA

sin=\frac{opposite}{adjacent}\\ cos=\frac{adjacent}{hypotenuse} \\tan=\frac{opposite}{adjacent}

23.

Find the missing side using Tangent

tan(30)=\frac{x}{4}

4*tan(30)=x

\frac{4\sqrt{3} }{3}

24.

Find the missing side using Tangent

tan(30)=\frac{x}{7}

7*tan(30)=x

\frac{7\sqrt{3}}{3}

25.

Find the missing side using Tangent

tan(30)=\frac{x}{18}

18*tan(30)=x

6\sqrt{3}

26.

Find the missing side using Tangent

tan(30)=\frac{x}{10}

10*tan(30)=x

\frac{10\sqrt{3} }{3}

27.

Find the missing side using Tangent

tan(30)=\frac{x}{14\sqrt{3} }

(14\sqrt{3}) *tan(30)=x

14

28.

Find the missing side using Tangent

tan(30)=\frac{x}{12}

12*tan(30)=x

4\sqrt{3}

5 0
3 years ago
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