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Free_Kalibri [48]
3 years ago
15

If D is the midpoint of the segment AC and C is the midpoint of segment DB , what is the length of the segment AB , if AC = 3 cm

.
Mathematics
2 answers:
Sergio039 [100]3 years ago
7 0
Only in middle school tried but cant do it

Sedaia [141]3 years ago
6 0

Answer:

<em><u>4.5 CM</u></em>

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❂✨Answered By Poopyeet✨❂

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Answer:

Final velocity is 3

Change in velocity is 4

Initial velocity is 2

Instantaneous velocity is 1

Acceleration is 5

7 0
3 years ago
10x is greater than or equal to 120 cm^2
Agata [3.3K]

Answer:

x greater than or equal to 12cm^2

Step-by-step explanation:

8 0
3 years ago
The equation tan(x- pi/3) is equal to _____.
kherson [118]

Answer:

B

Step-by-step explanation:

Using the expansion

tan(x - y) = \frac{tanx-tany}{1+tanxtany}, then

tan(x - \frac{\pi }{3})

=  \frac{tanx-tan(-\frac{\pi }{3}) }{1+tanxtan(-\frac{\pi }{3}) }

note that tan( -\frac{\pi }{3}) = - tan(\frac{\pi }{3}) = - \sqrt{3}

= \frac{tanx-(-\sqrt{3}) }{1+tanx(-\sqrt{3}) }

= \frac{tanx+\sqrt{3} }{1-\sqrt{3}tanx } → B

6 0
3 years ago
Read 2 more answers
figure below is made of rectangles what is the perimeter of the figure below note that not all the side lengths are labeled.
Ira Lisetskai [31]
I would love to help but there is no figure.
5 0
3 years ago
When Alice spends the day with the babysitter, there is a 0.6 probability that she turns on the TV and watches a show. Her littl
lara [203]

Answer:

A. P("Both Alice and Betty watch TV") = 12/25

B. P("Betty watches TV") = 12/25

C. P("Only Alice watches TV") = 3/25

Step-by-step explanation:

A. Because what we are told, Betty need that Alice turn on the TV, so, we first need the probability that she watches TV:

P(Alice)=3/5 (0.6)

And we know that:

P(Betty)=4/5 (0.8)

If we want the probability of both things happening at the same time (If you use a tree diagram, those events will be in the same branch), we proceed multiply them:

P("Both Alice and Betty watch TV") = 3/5 * 4/5 = 12/25

And this is the answer

B. Considering that Betty needs Alice to turn on the TV, the probability of Betty watching TV is the same as if she is with Alice.

P("Betty watches TV") = P("Both Alice and Betty watch TV") = 12/25

C. We use the same process as part A, but with a little difference. We now multiply for the probability that Betty does not watch TV (Because they still be in the same branch).

P("Betty does not watch TV") = 1 - P("Betty")

P("Betty does not watch TV") = 1 - 4/5

P("Betty does not watch TV") = 1/5

And the answer for part C is:

P("Alice watches TV without Betty") = 3/5 * 1/5 = 3/25

5 0
3 years ago
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