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2 years ago

Sec6A - tan6A = 1+3tan2A.sec2A

Mathematics

1 answer:

larisa [96]2 years ago

6 0

sec ^6 A−tan ^6A=(1+tan ^2A−tan ^2 A)(1+tan ^4

A+2tan ^2 A+tan ^2 A+tan ^4 A+tan ^4 A)

sec ^6 A−tan ^6 A=1+3tan ^2 A+3tan ^4A

L.H.S. = R.H.S.

Hence proved

Step-by-step explanation:

♡´･ᴗ･`♡

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Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and

kobusy [5.1K]

Answer:

a) $E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+$

b) $E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

c) $E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

d) $E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

e) $E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

f) $E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

g) $E(u) = E(v) +E(X) = 0+1=1$

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

$Y = 1+X +u$

$u = v+X$

with both Y and u random variables, we also know that:

$[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2$

And we want to calculate this:

Part a

$E(u|X=1)= E(v+X|X=1)$

Using properties for the conditional expected value we have this:

$E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1$

Because we assume that v and X are independent

Part b

$E(Y| X=1) = E(1+X+u|X=1)$

If we distribute the expected value we got:

$E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

Part c

$E(u|X=2)= E(v+X|X=2)$

Using properties for the conditional expected value we have this:

$E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

Because we assume that v and X are independent

Part d

$E(Y| X=2) = E(1+X+u|X=2)$

If we distribute the expected value we got:

$E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

Part e

$E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

Part f

$E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

Part g

$E(u) = E(v) +E(X) = 0+1=1$

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

8 0

3 years ago

349 times 3 expanded form

kozerog [31]

Answer:

1000 + 40 + 7

Step-by-step explanation:

MULTIPLY:

349 times 3 = 1,047

EXPANDED FORM:

1000 + 40 + 7

7 0

3 years ago

Which of the following graphs could be the graph of the function f(x)=-0.08x(x^2-11x+18)?

MariettaO [177]

Answer:

option C

Step-by-step explanation:

f(x)=-0.08x(x^2-11x+18)

Factor the parenthesis and find out the x intercepts

(x^2-11x+18)

We find out two factors whose product is 18 and sum is -11

-9 and -2 gives us product 18

-9 -2 gives us sum -11

(x^2-11x+18) =(x-9)(x-2)

f(x)=-0.08x(x^2-11x+18)

f(x)= -0.08x(x-9)(x-2)

set f(x) =0 and solve for x

-0.08x(x-9)(x-2) =0

-0.08x =0, so x=0

x-9 = 0 , so x=9

x-2 =0 , so x=2

x intercepts are (0,0) (2,0) and (9,0)

x intercepts are not repeating

so the graph crosses x -axis at x=0, 2,9

option C is correct

The graph is attached below4

5 0

3 years ago

Read 2 more answers

Write a division expression with a quotient of 20^{5}

ivolga24 [154]

Answer:

3200000

Step-by-step explanation:

6 0

3 years ago

On square PQRS below, if Q is located at (7,0) and R is located at (5,-8) what is the length of SR

jenyasd209 [6]

the length of SR is $2\sqrt{17}$ .

<u>Step-by-step explanation:</u>

Here we have , On square PQRS below, if Q is located at (7,0) and R is located at (5,-8) . We need to find Length of side SR . Let's find out:

We know that , In a square there are 4 sides , and length of every side is same . So side length of SR = side length of QR . Now , Let's find side length of QR .

We know that distance between two points $Q(x_1,y_1),R(x_2,y_2)$ given by :

⇒ $QR = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Two points given are $Q(7,0), R(5,-8)$ :

⇒ $QR = \sqrt{5-7)^2+(-8-0)^2}$

⇒ $QR = \sqrt{(2)^2+(-8)^2}$

⇒ $QR = \sqrt{4+64}$

⇒ $QR = \sqrt{68}$

⇒ $QR = 2\sqrt{17}$

But QR=SR , Therefore , the length of SR is $2\sqrt{17}$ .

8 0

4 years ago