Which of the following is an example of a discrete variable ?

Tacoma's population in 2000 was about 200 thousand, and has been growing by about 8% each year. If this continues, what will Tac

Brums [2.3K]

Answer: The population will be 408,000 people.

Step-by-step explanation:

So in 2000 there were 200,000 people and it started to grow 8% every year so up to 2013.

so find 8% of 200,000 and then multiply it by the the number of years.

8% * 200,000 = 16,000

Find the difference between the years.

2013 - 2000 = 13 years

13 * 16000 = 208000 This is the amount of new people from 2000 to 2013 so add it to the original population.

208,000 + 200,000 = 408,000

5 0

3 years ago

Henry baked 4 batches of 12

garri49 [273]

Answer: 43

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Consider a circle that has a circumference of 28/3.14 centimeters. What is the area of this circle? Answer fast plz also can you

bija089 [108]

Answer:

$A=6.33\ cm^2$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The circumference of a circle is equal to

$C=2\pi r$

we have

$C=\frac{28}{3.14}\ cm$

$\pi=3.14$

substitute the values

$\frac{28}{3.14}=2(3.14)r$

$r=\frac{28}{19.7192}\ cm$

$r=1.42\ cm$

step 2

Find the area of the circle

The area of the circle is equal to

$A=\pi r^{2}$

substitute the values

$A=(3.14)(1.42)^{2}$

$A=6.33\ cm^2$

7 0

3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

List all the perfect cubes up to 1000 and their cube roots.

DiKsa [7]

Hello from MrBillDoesMath!

Answer:

1^3 = 1

2^3 = 8

3^3 = 27

4^3 = 64

5^3 = 125

6^3 = 216

7^3 = 343

8^3 = 512

9^3 = 729

10^3 = 1000

Thank you,

MrB

4 0

3 years ago