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3 years ago

An orthocenter is the intersection of three.....?

Mathematics

2 answers:

Vsevolod [243]3 years ago

6 0

Answer: An orthocenter is the intersection of three: Altitudes in a triangle.

Step-by-step explanation:

Altitude--

An altitude is a line segment that originates from a vertex and is perpendicular to the line segment opposite to that vertex.

Orthocenter--

An orthocenter of a triangle is a point where the three altitudes of a triangle meet or intersect.In a right angled triangle the orthocenter is the vertex of the triangle where the right angle is formed.

steposvetlana [31]3 years ago

5 0

Answer:

i don't get it

Step-by-step explanation:

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X>0, y>0, 2x+3y=8, smallest value of xy? pls help me

torisob [31]

Answer:

where there is x in the equation we put 0

For y

=2(0)+3y=8

=0+3y=8 Group likely terms

=3y=8-0

=3y=8 Divide both sides by 3

=3y/3=8/3

Therefore y=2.6

For x

=2x+3y=8

=2x+3(0)=8

=2x+0=8 Group likely terms

=2x=8-0

=2x=8 Divide both sides by 2

=2x/2=8/2

Therefore x=4

The smallest numbers for x and y is 4 and 2.6 respectively

7 0

3 years ago

The lopez family goes to the movies. They buy 2 adult tickets for $6 each and 3 child tickets for $4 each. Write an equation to

andre [41]

I think it would be 2x+3y=$24
Adult tickets would be x
Child tickets would be y
I’m not 100% sure though sorry

3 0

3 years ago

Write the inequality represents the verbal expression-<br> All real numbers less than 69

Lorico [155]

Let be M the set of the real numbers less thant 69

so we write it as follow M = { x∈R / x ∠ 69}

7 0

3 years ago

Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0

4 years ago

Among all rectangles that have a perimeter of 176, find the dimensions of the one whose area is largest. write your answers as f

maria [59]

To get the largest area, the length and the width have to be as close as possible, if not the same.

Given Perimeter = 176 unit.
Length = 176 ÷ 4 = 44 units

Area = Length x Width
Area = 44 x 44 = 1936 units²

1936 units² is the largest possible area, given that the perimeter is 176 units. The dimension being 44 by 44

8 0

3 years ago

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