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3 years ago

A triangular field has sides of lengths 21, 39, and 41 yd. Find the largest angle. (Round your answer to the nearest degree.)

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

5 0

Answer:

80 degrees

Step-by-step explanation:

To start, use the fact that the largest angle is the opposite of the longest side. Now, use the Law of Cosines (a^2 = b^2 + c^2 - 2bc cos A), where a is the longest side and A is its opposite, and b and c are the other sides. Plug the values in to get 1681 = 1962 - 1638 cos A, or cos A = 281 / 1638. Now, use a calculator to get that A is approximately 80.12 degrees, or 80 degrees.

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6th grade math i mark as brainliest

Advocard [28]

Step-by-step explanation:

Use PEMDAS

$\begin{array}{ccc}214.8-(4^2+3)\times4&\\214.8-(16+3)\times4&Evaluate\\214.8-19\times4&Add\\214.8-76&Multiply\\138.8&Subtract\end{array}$

6 0

3 years ago

Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.

Vinvika [58]

Answer:

1) With initial centroids 10-40, final clusters are

first cluster (6,12,18,24,30)

second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

the points will be clustered according to their distance to the centroids.
Then centroids are updated as the cluster means.
This process continues until clusters doesn't change anymore

1) initial centroids 10-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

new centroids 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

final centroids 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = $(18-6)^{2}$ + $(18-12)^{2}$ + $(18-18)^{2}$ + $(18-24)^{2}$ + $(18-30)^{2}$ + $(45-42)^{2}$ + $(45-48)^{2}$ =378

2)initial centroids 10-20

first cluster (6,12) mean:9

second cluster (18,24,30,42,48) mean:32.4

new centroids 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

new centroids 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

final centroids 15-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

Sum of Squared errors = $(15-6)^{2}$ + $(15-12)^{2}$ + $(15-18)^{2}$ + $(15-24)^{2}$ + $(40-30)^{2}$ + $(40-42)^{2}$ + $(40-48)^{2}$ =348

6 0

3 years ago

What is the equation of the line –3x – 2y = 30 in slope-intercept form?

leonid [27]

-3x-2y=30
-2y=3x+30
y=-3/2x-15

the answer is y=-3/2x-15

4 0

3 years ago

The Insurance Institute for Highway Safety publishes data on the total damage caused by compact automobiles in a series of contr

erastova [34]

Answer: $Min : $750\ ,\ Q_1= \$800\ ,\ Median : \$925\ ,\ Q_3=\$1050\ ,\ Max: \$1100$

Step-by-step explanation:

The five -number summary consists of five values :

Minimum value , First quartile $(Q_1)$ , Median , Third Quartile $(Q_3)$ , Maximum value.

Given : The Insurance Institute for Highway Safety publishes data on the total damage caused by compact automobiles in a series of controlled, low-speed collisions.

The following costs are for a sample of six cars:

$800, $750, $900, $950, $1100, $1050.

Arrange data in increasing order :

$750,$800, $900, $950, $1050, $1100

Minimum value = $750

Maximum value = $1100

Median = middle most term

Since , total observation is 6 (even) , so Median = Mean of two middle most values ($900 and $950).

i.e. Median $=\dfrac{900+950}{2}=\$925$

First quartile $(Q_1)$ = Median of lower half ($750,$800, $900)

= $800

, Third Quartile $(Q_3)$ = Median of upper half ($950, $1050, $1100)

= $1050

Hence, the five-number summary of the total damage suffered for this sample of cars will be :

$Min : $750\ ,\ Q_1= \$800\ ,\ Median : \$925\ ,\ Q_3=\$1050\ ,\ Max: \$1100$

3 0

3 years ago

Solve the following:

vlada-n [284]

Answer: x > 8

2x - 7 > 9

2x > 9+7

2x > 16

2x / 2 > 16 / 2

x > 8

7 0

3 years ago