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3 years ago

Help for both of them

Mathematics

1 answer:

Nastasia [14]3 years ago

4 0

Answer:

1. acute angle with 74°

2. acute angle with 55°

Step-by-step explanation:

any angles less than 90° is acute

any angles that are 90° are right angles

any angles over 90° above are obtuse

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for every one rotation a small gear takes, the larger gear makes 1/6 of a rotation. If the gear makes 10 complete rotations, how

Vika [28.1K]

Answer:

Step-by-step explanation:

1 and 2/3 rotations. The little gear full rotation is equal to 1/6 of the big gears rotation. So if the big gear rotates once then its equal to 6 of the small rotations the extra 4 of the 10 rotations then becomes 4/6 of the big gears rotation or if simplified further, 2/3. This is my way of solving these questions

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3 years ago

What property illustrated 5+7+9=5+9+7

motikmotik

Answer:Associative Properties

Step-by-step explanation:

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3 years ago

Read 2 more answers

Plz help fast!!!! will mark brainliest if correct

Marta_Voda [28]

Step-by-step explanation:

(g+f)(x) ie option B

-2+1 = -1

1-2 = -1

4-5 = - 1

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3 years ago

the scale on a map shows that 5 centimeters equals 2 kilometers what numbers of the centimeters on the map represents the actual

jek_recluse [69]

2 cm would represent .8 of a kilometer

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4 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

3 years ago