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3 years ago

Name the point of concurrency of the angle bisectors. a. A b. B c. C d. not shown

Mathematics

1 answer:

Oxana [17]3 years ago

3 0

D. not shown , ...............

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HELP ME!!!! ASAP!!!!! Thank you so much!!!!

likoan [24]

3/2x + 1/5 >= -1
3/2x >= - 6/5
x >= -12/15
x >= -4/5

-1/2x - 7/3 >= 5

-1/2x >= 22/3
x <= -44/3

Its D

5 0

3 years ago

Which measure of center best represents this set of data? Number of television sets in the home: 4, 4, 2, 4, 4, 1, 3, 4, 2, 4 *

nataly862011 [7]

Answer:

Mean

Step-by-step explanation:

Given

$Data: 4, 4, 2, 4, 4, 1, 3, 4, 2, 4$

Required

Best measure of center

The mean is considered the best measure of center when there are no outliers in the given data.

Outlier is when a particular data set is far away from other datasets.

For instance: 20 is an outlier in the following data sets: $4, 4, 2, 20, 4, 4, 1, 3, 4, 2, 4$

Also, the given data are not open-ended.

An open-ended data is as follows: $1, 2, 3, 4, 5\ or\ more$

Since we do not have any of the above conditions, then the mean is the best center to use.

7 0

2 years ago

What do I do steps? please help

skad [1K]

Area = Length x Width

so, 88 = (something) x 8
Can you work out the something using mental math?

8 0

2 years ago

Factor the expression using the GCF. 18-12

rodikova [14]

Answer:

6(3-2)

Step-by-step explanation:

Find the GCF of both numbers. The GCF is 6. Write it as 6( - )

Then, divide both numbers by 6. 18 divided by 6 is 3 and 12 divided by 6 is 2. Fill in the blanks with 3 and 2. A=6(3-2)

7 0

3 years ago

Find the area each sector. Do Not round. Part 1. NO LINKS!!<br><br>

sladkih [1.3K]

Answer:

$\textsf{Area of a sector (angle in degrees)}=\dfrac{\theta}{360 \textdegree}\pi r^2$

$\textsf{Area of a sector (angle in radians)}=\dfrac12r^2\theta$

17) Given:

$\theta$ = 240°
r = 16 ft

$\textsf{Area of a sector}=\dfrac{240}{360}\pi \cdot 16^2=\dfrac{512}{3}\pi \textsf{ ft}^2$

19) Given:

$\theta=\dfrac{3 \pi}{2}$
r = 14 cm

$\textsf{Area of a sector}=\dfrac12\cdot14^2 \cdot \dfrac{3\pi}{2}=147 \pi \textsf{ cm}^2$

21) Given:

$\theta=\dfrac{ \pi}{2}$
r = 10 mi

$\textsf{Area of a sector}=\dfrac12\cdot10^2 \cdot \dfrac{\pi}{2}=25 \pi \textsf{ mi}^2$

23) Given:

$\theta$ = 60°
r = 7 km

$\textsf{Area of a sector}=\dfrac{60}{360}\pi \cdot 7^2=\dfrac{49}{6}\pi \textsf{ km}^2$

3 0

2 years ago