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3 years ago

Simplify. Please select the best answer from the choices provided

Mathematics

2 answers:

ehidna [41]3 years ago

6 0

The answer is 3 because 1/3 of 3 is 1 and 1/3 of 9 is 3 and 3 times 1 is 3

Answer: 3

Snezhnost [94]3 years ago

4 0

$3^{\frac{1}{3}} \times 9^{\frac{1}{3}} =$

$= 3^{\frac{1}{3}} \times (3^2)^{\frac{1}{3}}$

$= 3^{\frac{1}{3}} \times 3^{\frac{2}{3}}$

$= 3^{\frac{1}{3} + \frac{2}{3}}$

$= 3^1$

$= 3$

Answer: C. 3

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What is -4/7(-13/2) yeah

Zolol [24]

The answer is 26/7 it may seem to small to be the answer but just simplified it

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3 years ago

Determine the center and radius of each circle.<br> (x – 6)2 + y2 – 8y = 0

rusak2 [61]

Answer:

Step-by-step explanation:

(x-6)² + y²-8y = 0

Put the equation into center-radius form.

complete the square

coefficient of the y term: -8

divide it in half: -4

square it: (-4)² = 4²

add 4² to both sides to complete the square and keep the equation balanced:

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Answer:

Step-by-step explanation:

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3 years ago

Given: △ABC, E∈ AB m∠ABC=m∠ACE AB=34, AC=20 Find: AE

LenaWriter [7]

Answer:

AE = 11.76 units

Step-by-step explanation:

For better understanding of the solution, see the attached figure :

Given : E ∈ AB, m∠ABC = m∠ACE, AB = 34 and AC = 20

To find AE :

In ΔABC and ΔACE,

m∠ABC = m∠ACE ( Given )

∠A = ∠A ( Common angle for both the triangles )

By AA postulate of similarity of triangles, ΔABC ~ ΔACE

So, proportion of the corresponding will be equal.

$\implies \frac{AC}{AB}=\frac{CE}{BC}=\frac{AE}{AC}\\\\\implies\frac{AC}{AB}=\frac{AE}{AC}\\\\\implies\frac{20}{34}=\frac{AE}{20}\\\\\implies AE=\frac{20\times 20}{34}\approx 11.76$

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3 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

3 years ago