In beijing in 2008 the womens 3,000 meter steeplechase became an olympic event what is the distence in kilometer

Select the shapes that are similar to shape A

vekshin1

Answer:

Orange and pink, I'm not too sure

4 0

2 years ago

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Help.........................<br><br>

garri49 [273]

<h3>Answer: Choice A</h3>

$x^2\left(\sqrt[4]{x^2}\right)$

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Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

$\sqrt[4]{x} = x^{1/4}$

The same applies to x^10 as well

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}$

Multiply the exponents 10 and 1/4 to get 10/4

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}$

$\sqrt[4]{x^{10}} = x^{10/4}$

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If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

$x^{m/n} = x^a\sqrt[n]{x^b}$

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

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The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

$x^{m/n} = x^a\sqrt[n]{x^b}$

turns into

$x^{10/4} = x^2\sqrt[4]{x^2}$

which means

$\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}$

7 0

3 years ago

If r=10 and s=31 find R. Round to the nearest tenth

Vsevolod [243]

Answer: option c.

Step-by-step explanation:

You need to remember the identity:

$tan\alpha=\frac{opposite}{adjacent}$

The inverse of the tangent function is arctangent. You need to use this to calculate the angle "R":

$\alpha =arctan(\frac{opposite}{adjacent})$

You know that you need to find the measure of "R" and $r=10$ (which is the opposite side) and $s=31$ (which is the adjacent side), you can sustitute values into $\alpha =arctan(\frac{opposite}{adjacent})$

Then, you get:

$R=arctan(\frac{10}{31})\\\\R=17.9\°$

3 0

3 years ago

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Simplify <br> 3 (2x+4y-2z)+7 (x+y-4z)

Dima020 [189]

6x+12y-6z+7x7y-28z= 13x+19y-34z

6 0

3 years ago

Determine whether the points (–3,–2) and (3,2) are in the solution set of the system of inequalities below. y ≤ ½x + 2 y < –2

lana [24]

Given:

The system of inequalities:

$y\leq \dfrac{1}{2}x+2$

$y$

To find:

Whether the points (–3,–2) and (3,2) are in the solution set of the given system of inequalities.

Solution:

A point is in the solution set of the given system of inequalities if it satisfies both inequalities.

Check for the point (-3,-2).

$-2\leq \dfrac{1}{2}(-3)+2$

$-2\leq -1.5+2$

$-2\leq 0.5$

This statement is true.

$-2$

This statement is also true.

Since the point (-3,-2) satisfies both inequalities, therefore (-3,-2) is in the solution set of the given system of inequalities.

Now, check for the point (3,2).

$2$

This statement is false because $2>-9$ .

Since the point (3,2) does not satisfy the second inequality, therefore (3,2) is not in the solution set of the given system of inequalities.

7 0

3 years ago