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2 years ago

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What is the length of the hypotenuse of triangle ABC? Round your answer to the nearest tenth. A) 2.5 m B) 4.3 m C) 5.5 m D) 18.2

5 m

1 answer:

Natasha2012 [34]2 years ago

7 0

Hypotenuse= $\sqrt{1.5^{2}+ 4^{2} }$
=4.27

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If m RDC = 120°, then m DAC=?

monitta

The correct answer to this question is 240

From the image i've seen, the RDC has an angle of 120. So the major arc DC (that passes A and B) has an angle of 240.

So this makes that the measure of the angle of DAC is 240 because of the angle formed by RDC which is 120.

7 0

2 years ago

Read 2 more answers

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

Help meeèeeeeeeeeeeee

mamaluj [8]

The answer is B your welcome

3 0

2 years ago

Which number line illustrates the problem below? x = -1 - (-2) -3

Alina [70]

The answers to that would be x=-2

5 0

2 years ago

Help!! I will mark Brainliest!! Please show work! :) Thank You.

SSSSS [86.1K]

Answer:

The center is (20, 0.05) radius = 9

Step-by-step explanation:

The center is (20, 0.05) radius = 9

from the general form of the equation:

(x - h)^2 + (y - k)^2 = r^2

we see that r^2 = 81

so r = 9

3 0

2 years ago