How many elements are in the set {whole numbers between 3 and 15}?

Answers

Answer 1
Answer: It is 11 elements :)

Related Questions

Antony has a net spendable income of $1,950 per month. What is the maximum recommended amount of money that he should budget for food for a month? Round to the nearest dollar
Cassandra’s ruler is 22 centimetres long. April’s ruler is 30 centimetres long. How many centimetres longer is April’s ruler
Find m<1, m<2, and m<3
Let point L be between M and N on MN. Given that MN=31, ML= h-15 and LN = 2h-8. Find LN
SOMEONE PLEASE HELP WHATS 9x8x4??

If b^2-4ac=0 determine the need or real solutions of the equation ax^2+bx+c=0

Answers

If  B² - 4 A C = 0 ,

then

Ax² + Bx + C = 0 has two real roots, and they are equal.

In other words,   Ax² + Bx + C  is a perfect square.

construct the graph of x^2+y^2=9 what would this graph look like and what are its coordinates please help?

Answers

Answer:

The coordinates (3,0), (-3,0), (0,3) ,and (0,-3) helps us to draw the graph of the circle. Below is the explanation

Step-by-step explanation:

Given:

   The equation of a circle is x^(2) +y^(2) =9.

To find:

    Graph the given equation.

Let's find the center and radius by comparing the given equation with the standard form.

   (x-h)^(2) +(y-k)^(2) =r^(2)

 

We can rewrite our equation as (x-0)^(2)+ (y-0)^(2)= 3^(2)

Now, compare them

(x-h)^(2) +(y-k)^(2) =r^(2)

(x-0)^(2)+ (y-0)^(2)= 3^(2)

Center (h,k)=(0,0)

Radius r=3

Now, draw the graph of the circle by placing the center and radius.

So, the coordinates which are 3 units exactly from the center (0,0) help us to find the graph of the circle. So, (0,3), (0,-3), (3,0) ,and (-3,0) would help us to graph the circle.

You can learn more:

brainly.com/question/2870743

It would looks like circle with radius of 3 and its center at (0,0).


A micrometer is one-millionth of a meter (0.000001): ten thousand times shorter than a centimeter (0.01). How many micrometer one is one edge of a centimeter cube?

Answers

Answer:

       1cm^3=0.000000000001mm^3

Step-by-step explanation:

Given :  0.000001 mm = 0.01 cm

To find : How many mm^3 is in 1 cm^3

Solution :

Step 1 : Given  0.000001 mm = 0.01 cm

Step 2: Cubing both side (0.000001mm)^3 = (0.01cm)^3

Step 3: Solve  (10^(-6))^3mm^3=(10^(-2))^3cm^3

                    \Rightarrow((10^(-6))^3)/((10^(-2))^3)mm^3=1cm^3

                    \Rightarrow(10^(-18))/(10^(-6))mm^3=1cm^3

                    \Rightarrow 10^(-18+6)mm^3=1cm^3

                    \Rightarrow 10^(-12)mm^3=1cm^3

                     \Rightarrow 0.000000000001mm^3=1cm^3

 Therefore,  1cm^3=0.000000000001mm^3

If a cube has an edge of one centimeter and one centimeter is ten thousand times longer than a micrometer, then the edge has ten thousand micrometers.

I hope that's what you meant :)

5 miles to 35 miles in a fraction ?

Answers

5/35
Simplify by dividing both sides by 5
1/7
Final Answer: 1/7

The formula for perimeter of a rectangle is given by p = 2l + 2w, where p = perimeter, l = length, and w = width. solve the formula for w.

Answers

Perimeter formula
p = 2l + 2w

I reverse left-right side
2l + 2w = p

Move 2l to the right
2l + 2w = p
2w = p - 2l

Move 2 to the right
2w = p - 2l
w = (p - 2l)/2

Summary
p = 2l + 2w
2l + 2w = p
2w = p - 2l
w = (p-2l)/2

Ramon earns $1,715 each month and pays $53.40 on electricity. To the nearest tenth of a percent, what percent of Ramon's earnings are spent on electricity each month?

Answers

Percent of earnings spent on electricity each month is 3.1 %

Solution:

Given that Ramon earns $1,715 each month and pays $53.40 on electricity

To find: Percent of earnings spent on electricity each month

From given information,

Ramon's monthly salary = $ 1715

Electricity Rent = $ 53.40

Finding percentage of earnings spent on electricity:

Percent of earnings spent on electricity = \frac{\text{ electricity rent}}{\text{ monthly salary}} * 100

Substituting the values we get,

\rightarrow (53.40)/(1715) * 100\n\n\rightarrow 0.031137 * 100 = 3.113 \approx 3.1

Thus Percent of earnings spent on electricity each month is 3.1 %