<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B3%7D%7B16%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \frac{3}{16}

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4 years ago

13

} " alt=" \sqrt{ \frac{3}{16} } " align="absmiddle" class="latex-formula">
Rational or irrational?

1 answer:

sweet-ann [11.9K]4 years ago

7 0

Hello!
When you have a non-exact root (irrational number)

$\sqrt{ \frac{3}{16} } \to \frac{ \sqrt{3} }{ \sqrt{16} } \to \boxed{\boxed{\frac{ \sqrt{3} }{4} }}\Longleftarrow(irrational\:number)\end{array}}\qquad\quad\checkmark$

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Given g(x) = x squared - x, find g (2/3)

lesantik [10]

Answer:

$\large\boxed{g\left(\dfrac{2}{3}\right)=-\dfrac{2}{9}}$

Step-by-step explanation:

$g(x)=x^2-x=x(x-1)\\\\g\left(\dfrac{2}{3}\right)\to\text{put }\ x=\dfrac{2}{3}\ \text{to the eqation of the function:}\\\\g\left(\dfrac{2}{3}\right)=\left(\dfrac{2}{3}\right)\left(\dfrac{2}{3}-1\right)=\left(\dfrac{2}{3}\right)\left(\dfrac{2}{3}-\dfrac{3}{3}\right)=\left(\dfrac{2}{3}\right)\left(-\dfrac{1}{3}\right)=-\dfrac{(2)(1)}{(3)(3)}=-\dfrac{2}{9}$

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3 years ago

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3 years ago

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What is the explicit formula for the arithmetic sequence –7.5, –9, –10.5, –12...

natita [175]

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= - 1.5n - 6

Step-by-step explanation:

Given that the sequence is arithmetic with n th term ( explicit formula

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Here a₁ = - 7.5 and d = - 9 - (- 7.5) = - 9 + 7.5 = - 1.5

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Step-by-step explanation:

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3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

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Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

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4 years ago

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Step-by-step explanation:

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3 years ago