All categories

3 years ago

The segments shown below could form a triangle.

Mathematics

1 answer:

Shalnov [3]3 years ago

8 0

Answer:

Step-by-step explanation:

Condition for a triangle formed by three segments,

Sum of two segments should be greater than third segment

1). AC + BC > AB

2). AC + AB > BC

3). AB + BC > AC

1). AC + BC > AB

1 + 8 > 8

9 > 8

Triangle is possible.

2). AC + AB > BC

1 + 8 > 8

9 > 8

Triangle is possible.

3). AB + BC > AC

8 + 8 > 1

16 > 1

Triangle is possible.

Therefore, with the help of given segments triangle can be formed.

Answer is TRUE.

You might be interested in

Classify the following triangle. Check all that apply.

aksik [14]

B. acute and F. Scalene

6 0

3 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

3 years ago

In an examination , ten students scored the following marks : 60 , 58 , 90 , 51 , 47 , 81 , 70 , 95 , 87 , 99.

Feliz [49]

Answer:

Range = highest data - lowest data

99-47=

5 0

3 years ago

Plsss help meee!!!!!!

Lina20 [59]

Answer:

given below

Step-by-step explanation:

total area = rectangle area + 2 triangle area

= length * width = 2 ( $\frac{1}{2}$ * base * height )

= 7.5 * 6 + 2 ( $\frac{1}{2}$ * 3 * 4 )

= 57 ft²

separate it into two rectangles:

length * width + length * width

6 * 2 + 9 * 4

48 cm²

parallelogram area = base * height

= 20 * 25

= 500 in²

6 0

3 years ago

The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is dra

olga2289 [7]

Using the normal distribution, there is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For this problem, these parameters are given as follows:

$\mu = 180, \sigma = 20, n = 40, s = \frac{20}{\sqrt{40}} = 3.1623$

A sum of 7100 is equivalent to a sample mean of 7100/40 = 177.5, which means that the probability is the <u>p-value of Z when X = 177.5</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{177.5 - 180}{3.1623}$

Z = -0.79

Z = -0.79 has a p-value of 0.2148.

There is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.

More can be learned about the normal distribution at brainly.com/question/28135235

#SPJ1

5 0

2 years ago