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3 years ago

Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation

Mathematics

1 answer:

lawyer [7]3 years ago

6 0

Answer:

srry

Step-by-step explanation:

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What is the equation of a circle with center (1, -4) and radius 2?

kvv77 [185]

Answer:

(x-1)^2 + (y+4)^2 = 4

Step-by-step explanation:

The equation for a circle is given by

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x-1)^2 + (y- -4)^2 = 2^2

(x-1)^2 + (y+4)^2 = 4

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2 years ago

33. Due to some error in a weighing scale, it shows the following reading when at rest (no weight is placed on it). Neerja measu

Komok [63]

Answer:

The traditional scale consists of two plates or bowls suspended at equal distances from a fulcrum. One plate holds an object of unknown mass (or weight), while known masses are added to the other plate until static equilibrium is achieved and the plates level off, which happens when the masses on the two plates are equal. The perfect scale rests at neutral. A

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2 years ago

a farmer examinea sample of 25 egg cartons and finds three cracked eggs whast the best predictions of the number of cartons with

Lana71 [14]

It's C because 500/25 = 20 and 20x3=60

7 0

3 years ago

Can I have some help?

lisabon 2012 [21]

Answer:

$x = \sqrt{58}$

Step-by-step explanation:

first, you should draw a line to separate the triangle from the rectangle

You can get the height if the triangle by saying 16-9

This will give you 7

The base of the triangle is 3 because the opposite side is given and the opposite sides of a rectangle are equal

Now that we have the measurements of the triangle we can use the Pythagorean theorem

${c}^{2} = {a}^{2} + {b}^{2}$

${x}^{2} = {7}^{2} + {3}^{2}$

${x}^{2} = 49+ 9$

${x}^{2} = 58$

$x = \sqrt{58}$

3 0

2 years ago

Suppose that three numbers are selected one by one, at random, and without replacement, from the set of integers {1, 2, 3, . . .

fiasKO [112]

A - the third number falls between the first two

B - the first number is smaller than the second

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=n(n-1)(n-2)\\\\ |B|=\dfrac{n!}{2}\\\\ |A\cap B|=\dfrac{\dfrac{n!}{2}}{3}=\dfrac{n!}{6}\\\\ P(A\cap B)=\dfrac{\dfrac{n!}{6}}{n(n-1)(n-2)}=\dfrac{n!}{6n(n-1)(n-2)}=\dfrac{(n-3)!}{6}\\ P(B)=\dfrac{\dfrac{n!}{2}}{n(n-1)(n-2)}=\dfrac{n!}{2n(n-1)(n-2)}=\dfrac{(n-3)!}{2}\\\\ P(A|B)=\dfrac{\dfrac{(n-3)!}{6}}{\dfrac{(n-3)!}{2}}\\\\ P(A|B)=\dfrac{(n-3)!}{6}\cdot\dfrac{2}{(n-3)!}=\dfrac{1}{6}=\dfrac{1}{3}$

3 0

3 years ago