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AURORKA [14]
3 years ago
12

If the line segment at right is subdivided

Mathematics
1 answer:
Stells [14]3 years ago
8 0

Answer:

50

Step-by-step explanation:

75 dived into 3 parts can also be written as 75/3. 75/3 is 25 making each part worth 25. 2 segments are included in the marked portion so its 50 or (25 + 25)

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How to find vertical asymptotes and horizontal asymptotes
Alina [70]

Answer:

Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

Step-by-step explanation: Correct me if i'm wrong lol.

7 0
3 years ago
Es posible hacer un triángulo de 45 45 45 grados ?? <br><br> Por qué ?
nordsb [41]

Sí, porque un triángulo equilátero tiene 3 lados congruentes y 3 ángulos congruentes.

7 0
3 years ago
Which is the correct classification of StartFraction 3 Over 8 EndFraction?
Finger [1]

Answer:

Classifications would include

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Those would be classifications

4 0
3 years ago
Will award brainliest if correct
harkovskaia [24]

for the area of the smaller semicircles

area = 1/2 x 3.142 x 5 x 5

area = 39.275cm²

for the larger circle

area = 1/2 x 3.142 x 10x 10

area= 157.1cm²

but since part of it is out and is replaced again

..total area of shaded part is 157.1cm² + 39.275cm² - 39.275cm²

area =157.1cm²

3 0
4 years ago
Using a pair of compasses and a ruler only,construct ∆ABC where
LenKa [72]

1.Drawn a straight line AB=7 cm with the help of ruler.

2.With the help of compass drawn an arc from A and at the point where it cuts AB from that point made another arc drawn an arc cutting the previous arc.

3.From A drawn a straight line joining the arc and extend it to M.

4.With the help of ruler measured 5 cm and mark it as AC.

5.Joined BC and we get the required triangle.

6.From C drawn an arc and make it cut on AC and BC and from the point it cuts AC and BC drawn arc cutting each other and extend a line from point C extend a line to the point point of intersection of two arc.

7.Similarly we do for A and the point where the two line intersect denoted as O.

8.Made a perpendicular from O on AB this perpendicular will be radius and taking O as centre we draw a circle this is our incircle.

9.And AN is our locus of points equidistant from two lines AB and AC.

We need to construct a circle inscribed in triangle that is incircle it can be done by making angle bisector of two sides the point where it intersect will be incentre. The centre of required circle.

The angle bisector is the locus where points are equidistant from two sides.

6 0
3 years ago
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