<img src="https://tex.z-dn.net/?f=0%2C6%20whole5%2F9" id="TexFormula1" title="0,6 whole5/9" alt="0,6 whole5/9" align="absmiddle"

Svet_ta [14]

3 years ago

class="latex-formula">

Mathematics

2 answers:

kirill115 [55]3 years ago

5 0

Answer:

That doesn't make any sense

Step-by-step explanation:

Aleks04 [339]3 years ago

5 0

Answer:

What does that mean

Step-by-step explanation:

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Given that a, b, and c are non-zero real numbers and a + b ≠ 0, solve for x. ax + bx - c = 0

klemol [59]

The answer to the selected problem is: that is a linear equation.

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Which inequalities are true? Check all that apply. Three-fifths < 0 Four-sixths greater-than one-half Nine-elevenths < 1 T

Kipish [7]

$\frac{3}{5} < 0$

$\frac{8}{16} > \frac{1}{2}\\\\$

Solution:

We have to find the inequalities that are true

Option 1

$\frac{3}{5} < 0\\\\0.6 < 0$

0.6 is not less than 0

Thus this inequality is not true

Option 2

$\frac{4}{6} > \frac{1}{2}\\\\0.667 > 0.5$

0.667 is greater than 0.5

Thus this inequality is true

Option 3

$\frac{9}{11} < 1\\\\0.818 < 1$

0.818 is less than 1

Thus this inequality is true

Option 4

$\frac{8}{16} > \frac{1}{2}\\\\$

0.5 = 0.5

Thus the above inequality is not true

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pishuonlain [190]

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3 years ago

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Is 3 1/2 a whole number

Alina [70]

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3 years ago

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Solve: <img src="https://tex.z-dn.net/?f=%5Cunderset%7Bx%5Crightarrow~3%7D%7B%5Clim%7D~%5Cdfrac%7B2x%5E2-18%7D%7Bx%5E2-3x

vodomira [7]

Hello, please consider the following.

$\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }$

Thank you

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3 years ago