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slega [8]
3 years ago
8

Help Asap will give brainliest if you start answering by 21 minutes

Mathematics
1 answer:
Elza [17]3 years ago
8 0

<u>Answer:</u>

f(x) = (2x + 1)(2x − 1)

Explanation:

As we can see from the above function f(x) = 4x^{2} − 1, we need to find x-intercept, i.e., the value of x where this function's value is zero.

So let us Equate the function with zero

4x^{2} - 1 = 0 => 4x^{2} = 1 => x^{2} = 1/4 => x = \sqrt{1/4}

Hence we get 2 values of x, i.e., +1/2, and -1/2.

So now we need to check out the functions which are zero at this values of x, so let us substitute values of x in each and every option and see the values of functions.

Option A - f(x) = (4x + 1)(4x − 1) = 3, -3

Option B - f(x) = (2x + 1)(2x − 1)  = 0

Option C - f(x) = 4(x2 + 1)  = 5

Option D - f(x) = 2(x2 − 1) = -3/2

Hence answer is Option B

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3 years ago
Find m&lt;2<br> See picture for full problem. Please and thank you!
aev [14]

Step-by-step explanation:

as the other triangle is Isosceles triangle which means it's two sides are equal and two angles are equal so

to find its 3rd angle apply angles sum property on it

according to which

2(68) + x = 180

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x = 180 - 136

x = 44°

also we know, the other triangle is right angled triangle

so, it's one angle is 90° and 2nd angle is 44° (Vertically opposite angle)

therefore to know measure of angle 2 apply angle sum property of triangle in it too.

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hope this answer helps you dear!

6 0
2 years ago
If A=(0,0) and B=9(2,5), what is the approximate length of AB
Serhud [2]

Answer:

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Step-by-step explanation:

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2 years ago
Show that if X is a geometric random variable with parameter p, then
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Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

\sum_{k=1}^{\infty} b^{k-1}

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

And with this we have the requiered proof.

And since b=1-p we have that:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

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shtirl [24]
C. p=75/r
......................
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