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2 years ago

2. Which number is equivalent to 15% of 80? A. 1.2 B. 5.3 C. 12 D. 53

Mathematics

2 answers:

Nikolay [14]2 years ago

8 0

Answer: the answer is C= 12 :))

Step-by-step explanation:

kap26 [50]2 years ago

7 0

Answer: C 12

Step-by-step explanation: just multiply 80 times 0.15 and you will get 12

and if you want to check your work divide 12 by 0.15 and you will get 80

hope this helps mark me brainliest if it helped

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3 years ago

Solve the equation for all real solutions in simplest form.<br> z^2 - 12z +9= -3

Mars2501 [29]

Answer:

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

Step-by-step explanation:

We need to solve the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$

First simplifying the equation:

$z^2 - 12z +9+3= -3+3\\z^2 - 12z +12= 0$

Now, we can solve the equation using quadratic formula:

$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

we have a = 1, b=-12 and c=12

Putting values in formula and finding values of x

$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\z=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\z=\frac{12\pm\sqrt{144-48}}{2}\\z=\frac{12\pm\sqrt{96}}{2}\\z=\frac{12\pm9.8}{2}\\z=\frac{12+9.8}{2}\:or\:z=\frac{12-9.8}{2}\\z=10.9\:or\:z=1.1$

So, we get value of z: z=10.9 or z=1.1

Solving the equation for all real solutions in simplest form.

$z^2 - 12z +9= -3$ we get $\mathbf{z=10.9\: or\: z=1.1}$

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2 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%

Naddik [55]

We have

$\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)$

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$

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1 year ago