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DanielleElmas [232]
4 years ago
5

Explain the difference between an equation, like x=5, and an inequality, like x_> 5.

Mathematics
1 answer:
andrew11 [14]4 years ago
7 0
X=5 is a definite answer and one solution. While X greater than OR equal to 5 means that it could be 5 AND anything above 5, which gives multiple of choices for solutions. (I believe this write but it’s been awhile since I did algebra stuff)
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Step-by-step explanation:

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3 years ago
PLZ HELP!!! BRAINLIEST AND POINTS!!!
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The answer is for this question is c

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So the answer is c

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Sophia wrote an equation to represent the revenue of a movie theater for one day. She let x represent the number of child ticket
Angelina_Jolie [31]
The total revenue that is obtained from selling the tickets is the sum of the revenue from selling the child tickets and the adult tickets. If x and y are the number of child and adult ticket, the equation that would represent the given is,
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4 0
3 years ago
The baseball stadium holds 23,564 people. 3/4 of the seats were sold. How many people were in the stadium
Dennis_Churaev [7]

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3 0
3 years ago
Read 2 more answers
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
4 years ago
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