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vova2212 [387]
3 years ago
11

In a right triangle ABC, CD is the angle bisector of the right angle C. Two lines DF and DE are parallel to the legs of the tria

ngle. Prove that DFCE is a square.
find the following
m∠CED = 180° − m∠ _____ = ___ by reason ________
m∠CFD = 180° − m∠ _____ = ___ by reason ________
Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0
<span>m∠CED = 180° − m∠DEB= 90 degrees, by reason corresponding angles of parallel lines are congruent
</span>m∠CFD = 180° − m∠AFD = 90 degrees by reason corresponding angles of parallel lines are congruent
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One zero of f(x)=x^3-2x^2-5x+6 is 1 what are other zeros of the function
elena-s [515]

Answer:

The other zeros are -2 and 3.

Step-by-step explanation:

As 1 is a zero then (x - 1) is a factor so we divide:

x - 1 )x^3-2x^2-5x+6(x^2-x-6

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Answer:

Option A.

Step-by-step explanation:

From the first dot plot the given data set for week 1 is

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Divide the data in 4 equal parts.

(62, 68, 68, 68, 69), (70, 70, 72, 72, 72), (75, 75, 76, 78, 78), (80, 80, 80, 89, 89)

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From the second dot plot the given data set for week 2 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 85, 86, 86, 86, 88, 88, 88, 88, 89, 89, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69, 70, 70), 72, (72, 72, 75, 75, 76, 78, 78), (80, 80, 80, 85, 86, 86, 86), 88, (88, 88, 88, 89, 89, 89, 89)

Range=Maximum-Minimum=89-62=27

Median=\frac{78+80}{2}=79

Mean=\frac{\sum x}{n}=\frac{2364}{30}=78.8

The range for Week 1 is equal to the range for Week 2.

The median for Week 2 is more than the median for Week 1.

The mean for Week 2 is more than the mean for Week 1.

Therefore, the correct option is A.

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3 years ago
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Step-by-step explanation:

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