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vova2212 [387]
3 years ago
11

Find the measure of ∠ECB. Circle A is intersected by line CD at points D and E and line CB at point B, forming angle ECB outside

of the circle, the measure of angle ECB is x plus 10 degrees, arc EB is 6x plus 6 degrees, and arc DB 146 degrees.
Mathematics
2 answers:
goldfiish [28.3K]3 years ago
6 0

Answer:

its 45

Step-by-step explanation:

lubasha [3.4K]3 years ago
3 0

Answer:

m\angle ECB=50^o

Step-by-step explanation:

we know that

The measure of the external angle is the semi-difference of the arches it covers

so

In this problem

m\angle ECB=\frac{1}{2}(arc\ EB-arc\ DB)

substitute the given values

(x+10)^o=\frac{1}{2}((6x+6)^o-146^o)

solve for x

2x+20=6x-140\\6x-2x=20+140\\4x=160\\x=40

Find the measure of ∠ECB

substitute the value of x

m\angle ECB=(40+10)=50^o

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3 years ago
Ryan flew from Wiley Post to Ponca City and back. Ryan maintained an average rate of 450 mph going to Ponca City and an average
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The answer is B, and here's why.  Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
               d             r            t
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Let me explain this table to you.  The distance is d, we don't know what it is, that's what we are actually looking for.  We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back.  Hence, the d in both spaces.  There he flew 450 mph, back he flew 400 mph.  If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back.  For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back.  See? Now, because the distance there = the distance back, we can set the rt in both equal to each other.  If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other.  450t = 400(1-t) and
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3 years ago
Someone please help me with this!
Sveta_85 [38]

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Answer : 360

8 0
2 years ago
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