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ludmilkaskok [199]
3 years ago
12

Can the side lengths of 8cm 3cm 11cm form a triangle

Mathematics
1 answer:
Naily [24]3 years ago
3 0
No

Explanation: it depends on what kind of triangle you are doing so,I am going to assume that it is a regular triangle.
to be honest I don’t think there is a way to be able to put this into a triangle
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use the numbers 8, 6, and 2 to create one operation to write an expression that includes one exponent and has a value of 8. use
riadik2000 [5.3K]

Answer:

Answer:

Step-by-step explanation:

1) One solution is to write, as our exponent:

2) Because this is special case of the Exponents Law, valid for every base ≠ 0.

3) Hence, including an exponent the numbers 6, 2 whose value is eight.

Therefore our expression is true

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
Pa help po ako math problem po
andrew11 [14]

Answer:

i got 21.42 (rounded to the nearest second decimal) if not my full answer was 21.42147412. sorry if its incorrect

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Convert 6 2/5 to a percent?<br><br>A) 6.4%<br>B) 64%<br>C) 640%<br>D) 6,400%​
777dan777 [17]

Answer:

C) 640%

Step-by-step explanation:

6 2/5

Convert it to an improper fraction -

6 2/5 = 32/5

Convert the improper fraction to a decimal -

32/5 = 6.4

Multiply 6.4 by 100

= 640%

Hope this helps :)

4 0
3 years ago
Read 2 more answers
How can I do this exercise? <img src="https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B12%7D%20%3D%20%5Cfrac%7Bx%20%2B%201%7D%7Bx%20%2B
matrenka [14]

Answer:

-8/7 =x

Step-by-step explanation:

-2         x+1

----- = -----------

12         x+2

Using cross products

-2 (x+2) = 12 (x+1)

Distribute

-2x -4 = 12x+12

Add 2x to each side

-2x+2x -4 = 12x+2x+12

-4 = 14x +12

Subtract 12 from each side

-4-12 = 14x

-16 = 14x

Divide each side by 14

-16/14 =14x/14

-8/7 =x

3 0
3 years ago
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