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Alina [70]
3 years ago
13

What is the area of a triangle whose sides are 12 6 and 20 meters long

Mathematics
1 answer:
aleksley [76]3 years ago
3 0
38 meters long is the answer
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Please help, I will give you brainliest!
V125BC [204]

Answer:

12 = 12 \\ 12 - 12 = 12 - 12 \\ 0 = 0

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2 years ago
Choose the correct Triangle Congruence Theorem
Svetlanka [38]

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ASA

Step-by-step explanation:

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2 years ago
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Is 5(x+1) the same as 5x+1? Explain your answer.
Morgarella [4.7K]
No
5*x= x (correct)
5*1=5 (incorrect)
When I say correct or incorrect it means that it does/doesn’t match 5x+1
6 0
3 years ago
How does f(x) = 3 ^x change over the interval x = 4 to x= 5
zloy xaker [14]

Answer:

f(x) = 3^x increases steadily on the interval [4,5].

Step-by-step explanation:

This exponential function f(x) = 3^x has a positive base (3) which is larger than 1.  Thus, this function continues to increase as x increases, including the case where x increases from 4 to 5.

6 0
3 years ago
The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}ordered position

Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

4 0
3 years ago
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