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Kamila [148]
3 years ago
5

Find element a13 of matrix A, if A=

Mathematics
1 answer:
astraxan [27]3 years ago
8 0

\bf \begin{bmatrix} 2&-3&\boxed{4}\\ 5&0&-6\\ 7&2&-8 \end{bmatrix}\impliedby a_{1,3}\textit{ row 1, column 3}

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To verify the identity tan (x1+x2+x3) = tan x1 + tan x2 + tan x3 - tan x1 tan x2 tan x3 / 1 - tan x1 tan x2 - tan x2 tan x3
andrezito [222]

Answer:

The answer is the last one

tan(x_{1}+x_{2}+x_{3})=\frac{tanx_{1}+tan(x_{2}+x_{3})}{1-tanx_{1}tan(x_{2}+x_{3})}

Step-by-step explanation:

∵ tan(x_{1}+x_{2}+x_{3}=\frac{tanx_{1}+tan(x_{2}+x_{3})}{1-tanx_{1}tan(x_{2}+x_{3})}

=\frac{tanx_{1}+\frac{tanx_{2}+tanx_{3}}{1-tanx_{2}tanx_{3}}  }{1-tanx_{1}(\frac{tanx_{2}+tanx_{3}}{1-tanx_{2}tanx_{3}})}

Multiply up and down by 1-tanx_{2}tanx_{3}

\frac{tanx_{1}(1-tanx_{2}tanx_{3})+tanx_{2}+tanx_{3}}{1-tanx_{2}tanx_{3}-tanx_{1}tanx_{2}-tanx_{1}tanx_{3}}

=\frac{tanx_{1}+tanx_{2}+tanx_{3}-tanx_{1}tanx_{2}tanx_{3}}{1-tanx_{1}tanx_{2}-tanx_{2}tanx_{3}-tanx_{1}tanx_{3}}

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3 years ago
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30 people liked Brand X. This is 20% of the people in a survey.
Yuki888 [10]
A) 150 people
b) 80%
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3 years ago
Write 240 minutes in hours
miv72 [106K]
240 minutes is equal to 4 hours.
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17,985 + -33,789. i really need this question answered
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Answer:

-15804

Step-by-step explanation:

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Jose substitutes a value for y in the equation 5y = 3y + 4
sesenic [268]

Answer:

Step-by-step explanation:

Equation

5y = 3y + 4

combine like terms 3y becomes negative when moved to the left side being that its positive on the right. 4 has no variable so it stands alone on the right side.

-3y + 5y = 4

After adding a low negative to a high positive number. Turns into 2y.

2y = 4

Divide 2 on both sides to remove the number from the variable on the left.

\frac{2}{2}y = \frac{4}{2}

2 is canceled on the left side which gives the solution for Y.

y = 2

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3 years ago
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