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2 years ago

Need help with this question

Mathematics

1 answer:

torisob [31]2 years ago

4 0

D because if it keep rotating it will never hit anything

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A bag contains 10 marbles: 2 are green, 3 are red, and 5 are blue. Boris chooses a marble at random, and without putting it back

Anni [7]

Answer: 47/90

Step-by-step explanation:

3/10+2/9=47/90 somehow ig

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1 year ago

What is the reciprocal of -2/3

ryzh [129]

3/-2

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3 years ago

Read 2 more answers

PLZ HELP ASAP!!!!! <img src="https://tex.z-dn.net/?f=" id="TexFormula1" title="" alt="" align="absmiddle" class="latex-formula">

AveGali [126]

The simplified expression of $$4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}$ is $4x^3$ .

Solution:

Given expression is $4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}$ .

To simplify the given expression.

$$\Rightarrow4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}$

Using exponent rule: $\frac{1}{a^{-m}}=a^m$

$$\Rightarrow4x \cdot x^{5}.x^{-3}$

Using another exponent rule: $a^m \cdot a^n=a^{m+n}$

$$\Rightarrow4x \cdot x^{5-3}$

$$\Rightarrow4x \cdot x^{2}$

Using exponent rule: $a^m \cdot a^n=a^{m+n}$

$$\Rightarrow4x^{1+2}$

$$\Rightarrow4x^3$

Hence the simplified expression of $$4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}$ is $4x^3$ .

8 0

2 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

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2 years ago

Name two pairs of adjacent angels and t

antoniya [11.8K]

Answer:

That's the tea sis

Step-by-step explanation:

8 0

3 years ago