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3 years ago

Can the sides of a triangle have lengths 1, 3, and 4? yes OR no

Mathematics

2 answers:

lakkis [162]3 years ago

7 0

Answer:

Step-by-step explanation:

The 2 side lengths should be the same

seraphim [82]3 years ago

5 0

Answer:

Step-by-step explanation:

In order for the sides to be able to create a triangle two of the given sides have to be greater than the third side once added together.

$a+b > c$
$1+3=4$
$4=4$

Which means your answer is "no" these lengths do not create a triangle for it does not pass the Triangle Inequality Theorem.

Hope this helps.

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Roger manages an ice cream stand. He keeps track of how many minutes each customer stands in line.

aalyn [17]

0.5 = 2
0.75 = 2
1.0 = 3
1.25 = 4
1.5 = 5
2.0 = 2
2.25= 1
2.5 = 1

You should always count the numbers you used to make sure the total matched the total number of data. In this case, there are 20 numbers, and the total is also 20.

5 0

3 years ago

Read 2 more answers

How to do improper fractions with 3.8 repeating

vivado [14]

3.88888=3.88888/1=38.8888888

so lets ssay our repeating decimal = a fraction x
x=3.888 is true
multiply by 10
10x=38.8888
now subtract fir frist eqation from second
10x-x=38.8888-3.888888
9x=35
divide both sides by 9
x=35/9
x=3 and 7/9

anser is 3.88=3 and 7/9

3 0

3 years ago

Joseph built a sandbox in his backyard that is 2 meters wide, 4 meters long, and 2 meters tall. What volume of sand does he need

Wittaler [7]

Answer: B) 16 cubic meters

Step-by-step explanation:

To find the volume of a rectangular prism like this sandbox, multiply the length, height, and width together. 2*2*4 = 16.

4 0

2 years ago

Read 2 more answers

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago

Please help

nalin [4]

Answer:

12870ways

Step-by-step explanation:

Combination has to do with selection

Total members in a tennis club = 15

number of men = 8

number of women = 7

Number of team consisting of women will be expressed as 15C7

15C7 = 15!/(15-7)!7!

15C7 = 15!/8!7!

15C7 = 15*14*13*12*11*10*9*8!/8!7!

15C7 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C7 = 15*14*13*12*11/56

15C7 = 6,435 ways

Number of team consisting of men will be expressed as 15C8

15C8 = 15!/8!7!

15C8 = 15*14*13*12*11*10*9*8!/8!7!

15C8 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C8 = 6,435 ways

Adding both

Total ways = 6,435 ways + 6,435 ways

Total ways = 12870ways

Hence the required number of ways is 12870ways

4 0

2 years ago