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Pani-rosa [81]
3 years ago
11

Isolate y^2 in (x+4)^2+y^2=22

Mathematics
2 answers:
Mekhanik [1.2K]3 years ago
7 0

Answer:

Correct

Step-by-step explanation:

DiKsa [7]3 years ago
3 0

Answer:

i just need points

Step-by-step explanation:

hope you find the anwser

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Identify the percent of change as an increase or a decrease Of 15 to 21
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Answer:

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Step-by-step explanation:

4 0
2 years ago
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Are the two ratios equivalent? 6:21 and 8:28 Explain.​
Leviafan [203]

Answer:

The two ratios are equivalent.

Step-by-step explanation:

Reduce both fractions:

6/21 = 2/7  [Take out 3 from both sides]

8/28 = 2/7  [Take out 4 from both sides]

2/7 = 2/7

8 0
2 years ago
A player tosses a die 6 times.If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the ex
swat32

Answer:

E(x)=-0.2101

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x_1*p(x_1)+x_2*p(x_2)

Where x_1 and x_2 are the values that the variable can take and p(x_1) and p(x_2) are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that x_1 is equal to 2 and x_2 is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

P(a)=nCa*p^a*(1-p)^{n-a}

Where nCa=\frac{n!}{a!(n-a)!}

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability  p(x_1) that a player get at least two times number 6, is calculated as:

p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633

On the other hand, the probability p(x_2) that a player don't get  at least two times number 6, is calculated as:

p(x_2)=1-p(x_1)=1-0.2633=0.7367

Finally, the expected value of the amount that the player wins is:

E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101

It means that he can expect to loses 0.2101 dollars.

3 0
3 years ago
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3 years ago
Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d
faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

\int\limits^2_0 \ \dfrac{32}{x^2 +4}  \ dx

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint:  \dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1})

k = 4

(b) Now, lets evaluate the same integral using power series.

f(x) = \dfrac{32}{x^2 +4}

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

Step-by-step explanation:

(a)

\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx

= 32 \int\limits^2_0 \dfrac{1}{x+4}\  dx

=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0

= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))

= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))

= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))

= 32 (\dfrac{\pi}{8}-0)

= 32 ( (\dfrac{\pi}{8}))

= 4 \pi

The value of k = 4

(b) \dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...  \ \ \ \ \ (Taylor\ \ Series)

\int\limits^2_0  \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx

S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx +  \dfrac{3}{2^3}\int\limits^2_0 x^4 dx -  \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...

S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...

S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2)  + \dfrac{3}{9}(2^2) -...

S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

6 0
3 years ago
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