Which is the same as multiplying a number by two?

HELP! What is Jacobson's conjecture?

Elden [556K]

In abstract algebra, Jacobson's conjecture is an open problem in ring theory concerning the intersection of powers of the Jacobson radical of a Noetherian ring.

8 0

2 years ago

How do you find theoretical probability?

shepuryov [24]

Step-by-step explanation:

<h2>Theoretical Probability Definition</h2>

Theoretical probability is the theory behind probability. To find the probability of an event using theoretical probability, it is not required to conduct an experiment. Instead of that, we should know about the situation to find the probability of an event occurring. The theoretical probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes.

<h3>Find the probability of rolling a 5 on a fair die</h3>

Solution: 

To find the probability of getting 5 while rolling a die, an experiment is not needed. We know that there are 6 possible outcomes when rolling a die. They are 1, 2, 3, 4, 5, 6.Therefore, the probability is,Probability of Event P(E) = No. of. Favourable outcomes/ No. of. Possible outcomes.P(E) = 1/6.Hence, the probability of getting 5 while rolling a fair die is 1/6.

I hope it helps

4 0

2 years ago

-4log(2x) = -8 if someone can help me with this problem i’ll venmo them 2$

Temka [501]

X=50
You can keep ur 2 dollars but I’m not explaining it

5 0

3 years ago

What is lim x→3 x^2 - x - 6/ x - 3

Kisachek [45]

<h3>Answer: D) 5</h3>

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Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, $\frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2$

Allowing us to say,

$\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\$

5 0

3 years ago

PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

8 0

2 years ago