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3 years ago

Choose the correct answer :

Mathematics

2 answers:

nikdorinn [45]3 years ago

7 0

The answer is 2 The answer is 2 The answer is 2

mylen [45]3 years ago

6 0

Answer:

Step-by-step explanation:

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34107cm

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almond37 [142]

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Two standard number cubes are thrown. Find the probability that the sum of the numbers showing is ten given that the first numbe

Sergio [31]

Given that the first number shows up as four. The second number should be six so that the sum would be ten. Therefore, its probability should be 1/6.

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3 years ago

If the ratio of a circle's sector to its total area is 1/3, what is the measure of

nalin [4]

Answer:

option A: 120°

Step-by-step explanation:

$Area \ of \ circle = \pi r^2\\\\Area \ of \ sector = \pi r^2 \frac{ \theta}{360}\\\\Ratio \ of \ sector \ to \ circle = \frac{1}{3}\\\\That is, \\\frac{1}{3} = \frac{\pi r^2 \frac{\theta}{360}} {\pi r^2}\\\\\\\frac{1}{3} = \frac{\theta}{360} \\\\\frac{360}{3} = \theta\\\\120 = \theta$

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3 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

6 0

3 years ago