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irina1246 [14]
4 years ago
9

A number k less than 10

Mathematics
1 answer:
svlad2 [7]4 years ago
5 0
Your answer would be

10 - k
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Find the rate of change
Kazeer [188]
The rate of change is 500.
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3 years ago
Evaluate the following expression. 2^-2<br> Answer needs to be a fraction
Shtirlitz [24]

Answer is.........: 1/4

7 0
3 years ago
Read 2 more answers
) Find the coordinates of the point A' which is the symmetric point
soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point  A(3, 2) with respect to the line 2x + y - 12 = 0.​

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

C( \dfrac{x+3}{2}, \dfrac{y+2}{2})

Also, product of slope will be -1 .( Since, they are parallel )

\dfrac{y-2}{x-3} \times -2  = -1\\\\2y - 4 = x - 3\\\\2y - x = 1

x = 2y - 1

So, C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})

Also, C satisfy given line :

2\times ( \dfrac{2y + 2}{2})  + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}

Also,

x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}

Therefore, the symmetric points isA'(\dfrac{31}{5}, \dfrac{18}{5}) .

7 0
3 years ago
which choice shows 13⋅(7⋅4) correctly rewritten using the associative property and then correctly simplified? 13⋅4⋅7
Cloud [144]

The answer choice which represents the correctly rewritten form of the expression is; (13⋅7)⋅4.

<h3>Which answer choices represents the expression according to the associative property?</h3>

It follows from the task content that tee expression given is; 13⋅(7⋅4).

However, it follows from the associative property of multiplication and addition is such that the equivalent expression can be rewritten as;

  • (13⋅7)⋅4.

Read more on associative property;

brainly.com/question/13181

#SPJ1

7 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
3 years ago
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