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2 years ago

PLEASE HELP!! Find the missing side and round answer to the nearest tenth.

Mathematics

1 answer:

Mariulka [41]2 years ago

4 0

I got 18. i did this by using the SOH CAH TOA method. In this case you have the the angle and the hypotenuse, and you’re looking for the adjacent side, therefore you need to use CAH

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PLEASE HELP, ASAP. I WILL MARK YOU BRAINLIEST‼️

Alex787 [66]

Im sorry but i really cant see any of the words or number and ik this

4 0

3 years ago

5(x-10)=30-15x Ox=1 Ox=4 Ox=5 Ox=8

makvit [3.9K]

Answer:

x = 4

Step-by-step explanation:

5x-50=30-15x

5x +15x =30+50

20x= 80

Divide both sides of the equation by 20

x = 4

I hope it helps

7 0

3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

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

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago

Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

azamat

Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.

Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(a - b) • (a - c) = 0

or

(a - b) • (b - c) = 0

or

(a - c) • (b - c) = 0

We have

a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)

a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)

b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)

Case 1: If (a - b) • (a - c) = 0, then

(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6

which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (a - b) • (b - c) = 0, then

(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2

which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (a - c) • (b - c) = 0, then

(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0

but the solutions to x here are non-real, so we throw out this case.

So there are two possible values of x that make a right triangle, x = -6 and x = -2.

3 0

3 years ago

Hurry! might mark brainliest! what is this in slope-intercept form?

Angelina_Jolie [31]

Y=2/3x

The y-intercept is zero, and the slope is 2/3

5 0

3 years ago