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dexar [7]
3 years ago
5

I need asap pleaassee

Mathematics
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

om

Step-by-step explanation:

1-1

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PLZ I NEED HELP ILL GIVE YOU BRAINLIEST IF YOU HELP MEE
den301095 [7]
In the box at the top, it would be 8.
The boxes that are side by side would be 85+40.
the box all the way at the bottom would be 125
6 0
3 years ago
Anton will be constructing a segment bisector with a compass and straightedge, while Maxim will be constructing an angle bisecto
Genrish500 [490]

The similarities are;

  • Compass and a straight edge required for both construction
  • Both construction includes a line drawn from the intersection of arcs to bisect a segment or an angle
  • The bases for the construction of both bisector are the ends of segment and the angle to be bisected
  • The width of the compass when drawing intersecting arcs, is more than half the width of the segment or angle being bisected

The differences are;

  • Two points of intersection of arcs are used in the segment bisector while only one is requited in an angle bisector
  • The bisecting line crosses the segment in a segment bisector, while it stops at the vertex of the angle being bisected in an angle bisector

The sources of the above equations are as follows;

The steps to construct a segment bisector are;

  • Place the needle of the compass at one of the ends of the line segment to be bisected
  • Widen the compass so as to extend more than half of the length of the segment to be bisected
  • Draw two arcs, one above, and the other below the line
  • Place the compass needle at the other end and with the same compass width draw arcs that intersects with the arcs drawn in the above step
  • Draw a line segment by placing the ruler on the points of intersection of the arcs above and below the line

The steps to construct an angle bisector are;

  • With the compass needle at the vertex, open the pencil end such that arcs can be drawn on the rays (lines) forming the angle
  • Draw an arc on both lines forming the angle
  • Place the compass needle at one of the intersection points and draw an arc in between the lines forming the angle
  • Repeat the above step with the same compass width from the other intersection point with the rays forming the angle
  • Join the point of intersection of the two arcs to the vertex of the angle to bisect the angle

Therefore, we have;

The similarities are;

  • A compass and a straight edge can be used for both construction
  • A straight line is drawn from the point of intersection of arcs to bisect the segment or the angle
  • The arcs are drawn from the ends of the segment or angle to be bisected
  • The width of the compass is more than half the width of the line or angle when drawing the arcs

The differences are;

  • In a segment bisector, the intersection point is above and below the line, while in an angle bisector only one pair of arcs are drawn to intersect above the line
  • The bisecting line passes through the segment being bisected, while the line stops at the vertex in an angle bisector

Learn more about the construction of segment and angle bisectors here;

brainly.com/question/17335869

brainly.com/question/12028523

7 0
2 years ago
What is fifty five hundredths plus twenty five hundredths
Maksim231197 [3]
That would be 8000.. hope it helps
6 0
3 years ago
Read 2 more answers
The cooking club made some pies to sell at a basketball game to raise money for the school dance. The parents association contri
-BARSIC- [3]

The answer is 12 pies. If you sell 60 pieces and there are 5 pieces in each pie then you would Have 12 pies. X = pies   5x =60   To undo this you would have to divide 60 by 5.

6 0
3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
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