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3 years ago

E Reasons Y...

Mathematics

1 answer:

Gnom [1K]3 years ago

3 0

Answer:

x = .5

Step-by-step explanation:

Since we have a right triangle, we can use trig functions

tan theta = opp / adj

tan 63 = 1/x

x tan 63 = 1

x = 1/ tan 63

x=0.50952

Rounding to the nearest tenth

x = .5

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Use inductive reasoning to find the next term in this sequence 2,3,5,9,17. Then explain the rule for the pattern.

NISA [10]

We have been given the sequence 2,3,5,9,17.

We can write the terms of this sequence as

$2=2^0+1\\
3=2^1+1\\
5=2^2+1\\
9=2^3+1\\
17=2^4+1\\$

From the above term we can see that for the first term we take exponent 0 on 2 and then add 1 .

For second term we take exponent 1 on 2 and then add 1 .

For third term we take exponent 2 on 2 and then add 1 .

Using this fact for the next term of the sequence i.e. 6th term, we can take exponent 5 on 2 and then add 1 .

Therefore, next term of the sequence is given by

$2^5+1\\
=32+1\\
=33$

Therefore, the next term is 33.

Using the above facts, the pattern is given by

$a_n=2^{n-1}+1$

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3 years ago

What is the value of x? Image included

Dominik [7]

Answer:

68 degrees

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Let f(x)=x2+15x+56

Ivan

Answer:

-8, -7

Step-by-step explanation:

f(x)=x^2+15x+56

0 =x^2+15x+56

What two numbers multiply together to make 56 and add together to 15

7*8 = 56

7+8=15

0 = (x+8) (x+7)

Then use the zero product property

x+8 =0 x+7 =0

x+8-8 =0-8 x+7-7 =0-7

x=-8 x=-7

6 0

3 years ago

Read 2 more answers

Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????

LenaWriter [7]

Answer:

$E[X^2]= \frac{2!}{2^1 1!}= 1$

$Var(X^2)= 3-(1)^2 =2$

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

And we have that the moment generating function can be write like this:

$\phi(t) = e^{\frac{t^2}{2}$

And we can write this as an infinite series like this:

$\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...$

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

And then we can find the $E[X^2]$

$E[X^2]= \frac{2!}{2^1 1!}= 1$

And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

$E[X^4]= \frac{4!}{2^2 2!}= 3$

And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

7 0

3 years ago

A sector with a central angle measure of 200 degrees has a radius of 9cm

Helen [10]

Answer:

B) $45\pi cm^{2}$

Step-by-step explanation:

$A=\frac{\pi r^{2}(central. Angle) }{360} =\frac{\pi (9)^{2}(200) }{360} =\frac{\pi (81)(200)}{360}=45\pi cm^{2}$

Hope this helps

3 0

2 years ago