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3 years ago

QUESTION : how are integers and absolute value used in real world situations? THANK YOU ;)

1 answer:

4 0

Integers can be used to describe temperature, currency, and sea level.

Examples:
- A drop of 30 degrees = -30
- 25 meters above sea level = 25
-A 30-dollar deposit = 30

With this said, absolute values can also be used in real world situations.

Absolute value can be applied when it comes to distance, such as how far a object is from you, whether it's in front or behind you.

You might be interested in

How much interest is earned on $8,000 invested at an annual rate of 12% for 1 year?

Ghella [55]

Answer:

96$

Step-by-step explanation:

P = $8,000

R = 12%

T = 1yr

I = PRT/100

I = (800×12×1)/100

I = 96$

8 0

3 years ago

JJack, Jill, and Bill each carried a 48-fluid ounce bucket full of water down the hill. By the time they reached the bottom, Jac

Advocard [28]

Answer:

Jill lost 1/3

Bill lost 1/6

and Jack lost 1/4

Make a common denominator - let's use 12

Jill lost 4/12

Bill lost 2/12

Jack lost 3/12

Add them together = 9/12

reduce by dividing by 3.

total of 3/4 spilt

48 divided into 4 parts then multiplied by 3 of those parts

48 / 4 * 3 = 36

Step-by-step explanation:

3 0

2 years ago

What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64<img src="https://tex.z-dn.net/?f=x%5E3"

salantis [7]

Answer:

(3 + 4x)(9 - 12x + 16x²) = 0

(4m - 1)(16m² + 4m + 1) = 0

Step-by-step explanation:

Here we have to solve the sum of two cubes which is $27 + 64x^{3}$

Now, the equation is $27 + 64x^{3} = 0$

⇒ 3³ + (4x)³ = 0

⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0

⇒ (3 + 4x)(9 - 12x + 16x²) = 0

So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0

Therefore, from the above two relation we can solve for x.

One root will be $- \frac{3}{4}$ and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.

Again, we have to solve the difference of two cubes which is $64m^{3} - 1$

Now, the equation is $64m^{3} - 1 = 0$

⇒ (4m)³ - 1³ = 0

⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0

⇒ (4m - 1)(16m² + 4m + 1) = 0

So, (4m - 1) = 0 or (16m² + 4m + 1) = 0

Therefore, from the above two relation we can solve for m.

One root will be $\frac{1}{4}$ and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation. (Answer)