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3 years ago

Is 5/6 greater than 6/6

Mathematics

2 answers:

Anika [276]3 years ago

3 0

5/6 is smaller than 6/6 cause 6/6 is =1 5/6 is smaller than 1 whole

WARRIOR [948]3 years ago

3 0

No, say u have two pies.
Both of them, you cut into 6 piecies.
If u ate 5/6 of one pie it would be 5 pieces. If u also at 6 pieces of the other pie would be the same as 6/6
The second pie, (6 pieces gone), would be greater because it has more pieces gone out of the total

Also, 6/6 is equal to a whole. 5/6 is just a lityle bit less.

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3 years ago

Multiply 3x^2 (5x^3)

tresset_1 [31]

Answer:

Step-by-step explanation:

Here's what the multiplication looks like when it's done horizontally:

(4x2 – 4x – 7)(x + 3)

(4x2 – 4x – 7)(x) + (4x2 – 4x – 7)(3)

4x2(x) – 4x(x) – 7(x) + 4x2(3) – 4x(3) – 7(3)

4x3 – 4x2 – 7x + 12x2 – 12x – 21

4x3 – 4x2 + 12x2 – 7x – 12x – 21

4x3 + 8x2 – 19x – 21

That was painful! Now I'll do it vertically:

4x^2 – 4x – 7 is positioned above x + 3; first row: +3 times –7 is –21, carried down below the +3; +3 times –4x is –12x, carried down below the x; +3 times 4x^2 is +12x^2, carried down to the left of the –12x; second row: x times –7 is –7x, carried down below the –12x; x times –4x is –4x^2, carried down below the +12x^2; x times 4x^2 is 4x^3, carried down to the left of the –4x^2; adding down: 4x^3 + (+12x^2) + (–4x^2) + (–12x) + (–7x) + (–21) = 4x^3 + 8x^2 – 19x – 21

That was a lot easier! But, by either method, the answer is the same:

4x3 + 8x2 – 19x – 21

Simplify (x + 2)(x3 + 3x2 + 4x – 17)

I'm just going to do this one vertically; horizontally is too much trouble.

Note that, since order doesn't matter for multiplication, I can still put the "x + 2" polynomial on the bottom for the vertical multiplication, just as I always put the smaller number on the bottom when I was doing regular vertical multiplication with just plain numbers back in grammar school.

x^3 + 3x^2 + 4x – 17 is positioned above x + 2; first row: +2 times –17 is –34, carried down below the +2; +2 times +4x is +8x, carried down below the x; +2 times 3x^2 is +6x^2, carried down to the left of the 8x; +2 times x^3 is +2x^3, carried down to the left of the +6x^2; second row: x times –17 is –17x, carried down below the +8x; x times +4x is +4x^2, carried down below the +6x^2; x times +3x^2 is +3x^3, carried down below the +2x^3; x times x^3 is x^4, carried down to the left of the +3x^3; adding down: x^4 + (+2x^3) + (3x^3) + (+6x^2) + (+4x^2) + (+8x) + (–17x) + (–34) = x^4 + 5x^3 + 10x^2 – 9x – 34

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3 years ago

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3 years ago

If points A and B are randomly placed on the circumference of a circle with a radius of 2 cm, what is the probability that the l

OLga [1]

Answer:

The probability that the length of chord AB > 2 cm is 2/3 ⇒ D

Step-by-step explanation:

* Lets explain how to solve the problem

- Points A and B are randomly placed on the circumference of a

circle with a radius of 2 cm

- We need to find the probability that the length of chord AB is

greater than 2 cm

- Probability = required length of arc / circumference

- Assume that the length of the chord AB is 2 and the center of

the circle is labeled M

- The radius of the circle is 2 and AB = 2

- In Δ AMB

∵ MA= MB = 2 radii

∵ AB = 2

∴ Δ AMB is an equilateral triangle

∴ m∠AMB = π/3

∵ The length of an arc = r Ф , where r is the radius of the circle and

Ф is the central angle subtended by this arc

∵ r = 2 , Ф = π/3

∴ The length of arc AB = 2 × π/3 = 2π/3

- The arc AB is in the half of the circle then there are another arc

with the same length in the other half of the circle

- The length of the arc we must excluded from the length of the

circle to have the part of length of AB is greater than 2 is

2(2π/3) = 4π/3

∴ The length of the excluded arc is 4π/3

∵ The length of the circle is 2πr

∵ r = 2

∴ The length of the circle = 2π(2) = 4π

- The required arc = length circle - length excluded arc

∵ The length of the excluded arc = 4π/3

∵ The length of the circle = 4π

∴ The required arc = 4π - 4π/3 = 8π/3

- Probability = required length of arc / circumference

∴ Probability = (8π/3)/(4π) = 2/3

∴ The probability that the length of chord AB > 2 cm is 2/3

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3 years ago

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pashok25 [27]

Answer:

23 square units

Step-by-step explanation:

I just counted them because the area is the inside or the shaded region but your supposed to find the x and y values of each box multiply them and add them all up to get the answer

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3 years ago

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