What is the range of the relation below?

How you know that the values of the digit in the numbers 1500,000 and 100,500 ate not the same

Sliva [168]

In this number 1500,000,

1 is at million place and 5 is at hundred thousand place

So, the value of 1 = 1 (1,000,000) = 1 million

And the value of 5 = 5 (100,000 ) = 500,000 = Five hundred thousand

In the second number 100,500

1 is at hundred thousand place and 5 is at hundred place

The value of 1 = 1 (100,000) = 100,000 = Hundred thousand

And the value of 5 = 5 (100 ) = 500 = Five hundred

Since the digits 1 and 5 both are at different places in both the numbers so 1,500,000 and 100,500 are not the same.

6 0

3 years ago

I am really in hurry can anyone help me please

eimsori [14]

Answer:

This is your answer ☺️☺️☺️

6 0

2 years ago

Haley placed 5 of the bookmarks end to end Draw a point on the number line below to represent the total length of the 5 bookmark

Liula [17]

<-------------.-------------->
0 1 2 3 4 5 6 7 8 9 10

5 0

3 years ago

Veterinarian needs to know an animal's weight in kilograms is 20 pounds is about 9 kilograms and a dog weighs 30 pounds use a ra

givi [52]

Pounds: 10 20 30
Kilos: 4.5 9 13.5

half of 20 is 10, and half of 9 is 4.5. So every 10 pounds is 4.5 kilos. Add 10 to 20 to get 30, and add 4.5 to 9 to get 13.5. Both parts of the ratio table are increasing at the same rate (either 10 eachtime or 4.5 each time).

8 0

2 years ago

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

Harrizon [31]

Answer:

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

Step-by-step explanation:

We need to Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

First we will find slope of the line given in graph

The slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have points (1,6) and (2,2)

Slope is:

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-6}{2-1}\\Slope=\frac{-4}{2}\\Slope=-2$

Part a)

Write an equation of the line that passes through the given point and is (a) parallel

When the lines are parallel, they have same slope. So, slope of required line: m = -2

Using slope m =-2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=-2(4)+b\\3=-8+b\\b=3+8\\b=11$

The equation of line will be:

$y=mx+b\\y=-2x+11$

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Part b)

Write an equation of the line that passes through the given point and is (b) perpendicular

When the lines are perpendicular they have opposite slope. So, slope of required line: m = 1/2

Using slope m =1/2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=\frac{1}{2}(4)+b\\3=2+b\\b=3-2\\b=1$

The equation of line will be:

$y=mx+b\\y=\frac{1}{2}x+1$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

6 0

2 years ago