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3 years ago

Use the number line to compare the numbers -1/2 , -1 9/10, 3/5, 0.4

Mathematics

1 answer:

Leto [7]3 years ago

7 0

Answer:

C. -1 9/10, -1/2, 0.4, 3/5

Step-by-step explanation:

Using a number line, we can see that negative numbers are to the left of '0' and positive numbers are to the right of '0'. To compare numbers for increasing order, we would start with the lowest number to the farthest left on the number line to the greatest number to the farthest right. Placing the following numbers on the number line, the furthest to the left is -1 9/10, then -1/2, followed by 0.4 and then the greatest, 3/5. You can also compare by converting them all into decimals:

-1 9/10 = -1.9

-1/2 = -0.5

0.4 = 0.4

3/5 = 0.6

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35. Explain the difference between the reciprocal of x^2

Scorpion4ik [409]

Answer:

As you can see, the difference between the reciprocal of $x^2$ and the inverse of $x^2$ is that $\frac{1}{x^2} =x^{-2}$ and $\sqrt{x} =x^{\frac{1}{2}$ .

Step-by-step explanation:

First lets find both the reciprocal of $x^2$ and the inverse of

Recall that the reciprocal of a value is where you take a fraction and swap the places of the terms. In the case of $x^2$ , 1 is the denominator, so

$\frac{x^2}{1} =\frac{1}{x^2}$

To find the inverse of a function, you first need swap the locations of x and y in the equation

$y=x^2\\\\x=y^2$

Now, you need to solve for y

$y^2=x\\\\y=\sqrt{x}$

Now, lets rewrite each of these to better compare them

$\frac{1}{x^2} =x^{-2}\\\\\sqrt{x} =x^{\frac{1}{2}$

As you can see, the difference between the reciprocal of $x^2$ and the inverse of

7 0

3 years ago

Hannah went to the fair with her friends. She spent $7 to enter the fair and decided to play games until she won a stuffed anima

Evgesh-ka [11]

Answer:

24 games

Step-by-step explanation:

7 to enter

0.5 per game

19 dollars spent

19 = 7 + 0.5x

19 - 7 = 0.5x

12 = 0.5x

x = 12/0.5

x = 24

5 0

3 years ago

What is the equation of the line that is parallel to y-3x=2 and that passes through (6,1)?

Rudik [331]

Slope-intercept form: y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be parallel, they need to have the same slope.

y - 3x = 2 Add 3x on both sides to change the equation to slope-intercept form

y - 3x + 3x = 2 + 3x

y = 3x + 2 The slope is 3, so the parallel line's slope is also 3.

Now that you know the slope, substitute/plug it into the equation

y = mx + b

y = 3x + b To find "b", plug in the point (6, 1) into the equation, then isolate/get the variable "b" by itself

1 = 3(6) + b

1 = 18 + b Subtract 18 on both sides to get "b" by itself

1 - 18 = 18 - 18 + b

-17 = b

y = 3x - 17 Your answer is the 1st option

8 0

3 years ago

Read 2 more answers

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago

Are all integers rational numbers?

Snezhnost [94]

Answer:

True

Step-by-step explanation:

Yes! All integers "are" rational numbers. They can be expressed in the form of p/q.

Hope this helps :D

Don't hesitate asking anything regarding this.

5 0

3 years ago

Read 2 more answers