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3 years ago

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]

Answer:

ASA

Step-by-step explanation:

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2 years ago

Read 2 more answers

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