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3 years ago

What is the place value relationship of the 9's in 299

Mathematics

2 answers:

eimsori [14]3 years ago

8 0

Ones and tens thats the anwers

Jobisdone [24]3 years ago

4 0

The ones place and tens place are nine.

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Solve for x. Leave your answer in simplest radical form.

pogonyaev

Answer:

x = 7

Step-by-step explanation:

I used the formula c^2 = a^2 + b^2

if you need further explanation,let me know

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2 years ago

Use what you know about sine, cosine, and tangent to solve for the missing angle.

Wittaler [7]

Answer:

it is a 30 degree angle

Step-by-step explanation:

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3 years ago

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X subtract 9 equals 3 / 5X

Korvikt [17]

Answer:

$x=22\frac{1}{2}=22.5$

Step-by-step explanation:

$x-9=\frac{3}{5}x$

First, lets get rid of the fraction. I did this by multiplying both sides by $5$ .

$5(x-9)=5(\frac{3}{5}x)\\5x-45=\frac{5}{1} *\frac{3}{5}x\\5x-45=\frac{15}{5}x\\5x-45=3x$

We want to isolate $x$ on one side of this equation. Let's put any values with $x$ on one side of the equation, and normal integers on the other.

$2x=45$

Divide both sides by $2$ .

$x=\frac{45}{2}$

If your teacher wants you to leave your final answer as an improper fraction, your final answer is this. If they want it to be a mixed number or decimal, your final answer will be:

$x=22\frac{1}{2}=22.5$

7 0

3 years ago

Is 2/9 closer to 1/2 or 1 or 0

SVEN [57.7K]

2/9=0.222
The answer would be 0.

8 0

4 years ago

Read 2 more answers

Kamal wrote the augmented matrix below to represent a system of equations.

Sergio039 [100]

Answer:

$\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]$

Step-by-step explanation:

Given [Missing from the question]

$\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]$

Required

$R_2 \to -3R_2$

This implies that, we form a new matrix where the second row of the new matrix is a product of -3 and the second row of the previous matrix.

So, we have:

$Initial =\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]$

$New =\left[\begin{array}{cccc}1&0&1&|-1\\-3*1&-3*3&-3*-1&|-3*-9\\3&2&0&|-2\end{array}\right]$

$New =\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]$

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3 years ago

Read 2 more answers