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soldi70 [24.7K]
3 years ago
12

The following statement are true about complex fraction,except?​

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
4 0

Step-by-step explanation:

A fraction (from Latin fractus, "broken") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. A common, vulgar, or simple fraction (examples: {\displaystyle {\tfrac {1}{2}}}{\tfrac {1}{2}} and {\displaystyle {\tfrac {17}{3}}}{\displaystyle {\tfrac {17}{3}}}) consists of a numerator displayed above a line (or before a slash), and a non-zero denominator, displayed below (or after) that line. Numerators and denominators are also used in fractions that are not common, including compound fractions, complex fractions, and mixed numerals.

Paraphin [41]3 years ago
3 0
You didn’t even put the choices...
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Damm [24]

Answer:

2/5 or 0.4

Step-by-step explanation:

y2-y1/ x2-x1

y2=2, y1=0

x2=5, x1=0

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3 years ago
Which equation best represents a trend line for the scatter plot? These are fractions BTW.
kenny6666 [7]
(1, 8) and (7,1)
slope = (8-1)/(1-7) = -7/6
y = mx + b
b = y - mx
b = 8 - (-7/6)(1)
b = 8 + 7/6
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equation
y = -7/6(x) + 55/6

answer
<span>y=−7/6 x+55/6 </span>
6 0
2 years ago
Read 2 more answers
Suppose given △ABD and △CBD.
patriot [66]

Answer:

The required result is proved with the help of angle bisector theorem.

Step-by-step explanation:

Given △ABD and △CBD, AE and CE are the angle bisectors. we have to prove that \frac{AD}{AB}=\frac{DC}{CB}

Angle bisector theorem states that an angle bisector of an angle of a Δ divides the opposite side in two segments that are proportional to the other two sides of triangle.

In ΔADB, AE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment AD to the line segment AB.

\frac{DE}{EB}=\frac{AD}{AB}   →  (1)

In ΔDCB, CE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment CD to the line segment CB.

\frac{DE}{EB}=\frac{CD}{CB}    →  (2)

From equation (1) and (2), we get

\frac{AD}{AB}=\frac{CD}{CB}

Hence Proved.

5 0
3 years ago
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RideAnS [48]
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7 0
3 years ago
Consider an experiment with sample space S 5 50, 1, 2, 3, 4, 5, 6, 7, 8, 96 and the events A 5 {0, 2, 4, 6, 8} B 5 {1, 3, 5, 7,
Brut [27]

Answer with Step-by-step explanation:

S={0,1,2,3,4,5,6,7,8,9}

A={0,2,4,6,8}

B={1,3,5,7,9}

C={0,1,2,3,4}

D={5,6,7,8,9}

a.A'=S-A

A'={0,1,2,3,4,5,6,7,8,9}-{0,2,4,6,8}

A'={1,3,5,7,9}

b.C'=S-C

C'={0,1,2,3,4,5,6,7,8,9}-{0,1,2,3,4}

C'={5,6,7,8,9}

c.D'=S-D

D'={0,1,2,3,4,5,6,7,8,9}-{5,6,7,8,9}

D'={0,1,2,3,4}

d.A\cup B={0,2,4,6,8}\cup{1,3,5,7,9}

A\cup B={0,1,2,3,4,5,6,7,8,9}=S

e.A\cup C={0,2,4,6,8}\cup{0,1,2,3,4}

A\cup C={0,1,2,3,4,6,8}

f.A\cup D={0,2,4,6,8}\cup{5,6,7,8,9}

A\cup D={0,2,4,5,6,7,8,9}

8 0
3 years ago
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