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dlinn [17]
2 years ago
8

How to solve right triangles with no sides or angles given.

Mathematics
1 answer:
MrMuchimi2 years ago
8 0
Use The Law of Cosines to calculate the unknown side.
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How do I solve the multi step problem 8.5 - 1.2x = 6.7
LekaFEV [45]
First subtract 8.5 from both sides, then divide the -1.2 to get x by itself 


x=1.5
3 0
3 years ago
Only answer if you know! no spams :)
artcher [175]

Answer:

The correct answer is C.

Step-by-step explanation:

There are 8 slots on Spinner B.

3 of those slots have squares.

Hence out of all the slots, 3/8 of them have squares.

Hope this helps:) Goodluck!

5 0
2 years ago
Read 2 more answers
Thirteen more than the quotient of twenty and four
oksano4ka [1.4K]

Answer:

19

Step-by-step explanation:

20/4 + 13

20/4 = 6

6+13 = 19

8 0
3 years ago
Read 2 more answers
What does it mean when a number is out of the parenthesis? <br><br><br><br> Example 3($28+$32)-$10=
777dan777 [17]
When the number is out of parenthesis it means it is multiplying the number between the parenthesis 
example - 3 ( 28 + 32 ) => 3 ( 60)
which means 3 * 60 = 180

Note:- you cannot multiply the numbers when they are 3 ( 28 + 32 ) because you have to use PEMDAS ( order of operation ) to solve the problem where P = parenthesis , E = exponents, M = multiplication, D = divide , A = addition , S = subtract.
4 0
3 years ago
Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]

3 0
3 years ago
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