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3 years ago

Can you think of a situation where it would make the most sense to express your birth year in binary numbers?

Mathematics

1 answer:

telo118 [61]3 years ago

5 0

Answer:

Yeah I've thought of it and have an example with the year 2000 as the birth year

So 2000 In binary= 11111010000

Step-by-step explanation:

Binary numbers consist of only 0 and 1

So to convert a number to binary we divide by 2

Let's assume a birth year of 2000

Now let's convert it to binary

2000/2= 1000 R 0

1000/2= 500 R 0

500/2= 250. R 0

250/2= 125. R 0

125/2 = 62. R 1

62/2= 31. R 0

31/2= 15. R 1

15/2 = 7. R. 1

7/2= 3. R 1

3/2 = 1. R 1

1/2 = 0. R 1

Where R signifies the remainder

So we'd pick the remainder from the bottom and that becomes our binary number

So 2000 In binary= 11111010000

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$64.50 x 4(quarters in a year) = $258.00 per year

$258.00 x 6 years = $1,548.00

$1,500.00 + $1,548.00 = $3,048.00 balance after six years

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3 years ago

An adult and 5 children rent a skates. it costs $4 per adult for skates and a total of $16

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Answer:

$2.20 per child

Step-by-step explanation:

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3 years ago

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3 years ago

Can the sides of a triangle have lengths 7, 7, and 1?

Alika [10]

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Step-by-step explanation:

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2 years ago

MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

Given functions:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the quadratic formula and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$

Therefore, the solutions are:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

6 0

2 years ago