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zepelin [54]
3 years ago
10

a group of students was asked about the number of emails they each sent that day the results are 0,0,1,2,3,4,5,6,6,7,8,8,8,9 i w

ill post the other half now​

Mathematics
2 answers:
Over [174]3 years ago
6 0

Answer:

D is the answer

Step-by-step explanation:

All you do is count the values from 0 to 3 which there are 5 of and only D shows that (I got the remaining answer choices from your profiles other question showing the remaining answers)

olasank [31]3 years ago
3 0

Answer:

D

Step-by-step explanation:

All you do is count the values from 0 to 3 which there are 5 of and only D shows that (I got the remaining answer choices from your profiles other question showing the remaining answers)

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stich3 [128]

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, 145–√5  is the area of a triangle.

I supposed here that [ABD] is the perimeter of ▲ ABD.

As  BD  is a bisector of  ∠ABC ,

ABBC=ADDC=97

Let  ∠B=2α

Then in isosceles  △DBC

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Thus  AB=18cosα

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∠A=π−3α

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[1]8=7cos2α−9cosαcos(3α)

cos(3α)=cos(α+2α)=cosαcos(2α)−sinαsin(2α)=cosα(2cos2α−1)−2cosαsin2α=cosα(4cos2α−3)

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8=cos2α(7−9(4cos2α−3))

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First root lead to  α=π4  and  ∠BDC=π−∠DBC−∠C=π−2α=π2 . In such case  ∠A=π−∠ABD−∠ADB=π4, and  △ABD  is isosceles with  AD=BD. As  △DBC  is also isosceles with  BD=DC=7,  AD=7≠9.

Thus first root  cos2α=12  cannot be chosen and we have to stick with the second root  cos2α=49. This gives  cosα=23  and  sinα=5√3.

The area of a triangle ABD=12h∗AD  where h  is the distance from  B  to  AC.

h=BCsinC=14cosαsinα

Area of  triangle ABD=145–√5

= 145–√5.

Incomplete question please read below for the proper question.

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, what is the area of triangle ABD?

Learn more about the Area of the triangle at

brainly.com/question/23945265

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