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2 years ago

POSSIBLE POINTS 10

Mathematics

1 answer:

Sergio039 [100]2 years ago

6 0

Answer:

Minimum: 8 Q1: 14 Median: 17.5 Q3: 30 Maximum: 32

Step-by-step explanation:

8, 9, 11, 14, 17, 23, 29,30, 32, 32

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PLS HELP ME FAST PLS !!!!!!!!!!!

erastova [34]

Answer:

The Answer is P = (-2.5 , 1 )

Step-by-step explanation:

Let, the point is P(x,y)

then, $x=\frac{m_{1}x_{2}+ m_{2}x_{1}}{m_{1}+m_{2}}$

=> $x=\frac{5*(-1)+ 3*(-5)}{5+3}$

∴ $x=-2.5$

Again, $y=\frac{m_{1}y_{2}+ m_{2}y_{1}}{m_{1}+m_{2}}$

=> $y=\frac{5*7+ 3*(-9)}{5+3}$

∴ $y=1$

Thus, P = (-2.5 , 1 )

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2 years ago

QUiCK whoever gives me the right answer gets brainliest

8_murik_8 [283]

Answer:

speed of R miles per hour

miles depends on time

thats why it's miler per hour

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3 years ago

A kiddie pool holds miniature plastic ducks in different colors. Sophia randomly picks up a duck and then replaces it. She repea

aliya0001 [1]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

The table shows the number of flights leaving an airport. How many flights leave in four days?

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

1 times what equals 16

what =16

2 times what equals 32

what=16

etc....

The answer is 16 because the table shows multiples of 16.

16, 32, 48,64, 80....

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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

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

6 0

2 years ago