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3 years ago

6

How many 12-digit decimal sequences are there that start and end with a sequence of at least two 3s?

1 answer:

stiks02 [169]3 years ago

6 0

Four positions of such a sequence are given to be 3s, and we have 10 choices for the remaining eight positions. So the number of such sequences is

$1^4\cdot10^8=10^8=100,000,000$

or 100 million.

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PLS HELP ILL GIVE BRAINLIEST

inysia [295]

<h3>Answer: 14x - 8</h3>

=======================================================

Explanation:

I'll use the quadratic formula to find the roots or x intercepts. This slight detour allows us to factor without having to use guess-and-check methods.

The equation is of the form ax^2+bx+c = 0

a = 12
b = -11
c = -5

This leads to...

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-11)\pm\sqrt{(-11)^2-4(12)(-5)}}{2(12)}\\\\x = \frac{11\pm\sqrt{361}}{24}\\\\x = \frac{11\pm19}{24}\\\\x = \frac{11+19}{24} \ \text{ or } \ x = \frac{11-19}{24}\\\\x = \frac{30}{24} \ \text{ or } \ x = \frac{-8}{24}\\\\x = \frac{5}{4} \ \text{ or } \ x = -\frac{1}{3}$

Now use those roots to form these steps

$x = \frac{5}{4} \ \text{ or } \ x = -\frac{1}{3}\\\\4x = 5 \ \text{ or } \ 3x = -1\\\\4x - 5 =0 \ \text{ or } \ 3x+1 = 0\\\\(4x-5)(3x+1) = 0$

Refer to the zero product property for more info.

Therefore, the original expression factors fully to (4x-5)(3x+1)

Use the FOIL rule to expand it out and you should get 12x^2-11x-5 again.

----------------------------------------------

We did that factoring so we could find the side lengths of the rectangle.

I'm using the fact that area = length*width

L = length = 4x-5
W = width = 3x+1

The order of length and width doesn't matter.

From here, we can then compute the perimeter of the rectangle

P = 2(L+W)

P = 2(4x-5+3x+1)

P = 2(7x-4)

P = 14x - 8

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