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3 years ago

What is the additive inverse of 3? a. -3 b. a+(-a)=0 c. 0 d. -a

Mathematics

2 answers:

finlep [7]3 years ago

8 0

Answer:

A. -3

Step-by-step explanation:

3 and -3 are additive inverses since 3 + (-3) = 0. -3 is the additive inverse of 3.

stellarik [79]3 years ago

8 0

Answer:

a. -3

Step-by-step explanation:

Two numbers are additive inverses if they add to give a sum of zero. 3 and -3 are additive inverses since 3 + (-3) = 0. So, -3 is the additive inverse of 3.

You might be interested in

1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the

Musya8 [376]

Answer:

A) 34.13%

B) 15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

$z=\frac{x-m}{s}$

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

$z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0$

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

= 0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

$P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587$

C) Between 80 and 120

$P( 80$

D) less than 80

$P( x < 80 ) = P( z$

E) Between 70 and 100

$P( 70$

F) More than 130

$P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013$

8 0

3 years ago

2. Solve the equation by completing the square. Show your work.

scoundrel [369]

Answer:

x = 5 or 25

Step-by-step explanation:

I think the method I am about to explain is slightly quicker and easier than the method in your question. This works for any 'complete the square' question.

We begin with x² - 30x = - 125.

First, we are going to factorise the left-hand side of the equation by dividing the 'b' value (-30) by 2 (you'll see why this works in a minute):

(x - 15)²

We want these brackets to multiply out to give x² - 30x, so that they equal the left-hand side of the equation. Unfortunately, if we multiply them out, we get:

(x - 15)(x - 15) =

x² - 30x + 225

There is an unwanted term (the + 225, from 15²)! We only want x² - 30x, so we have to remove this term by subtracting it from the left side of the equation. To do this, let's set up the original equation again:

(x² - 30x + 225) - 225 = - 125

Note: The reason why we don't have to subtract it from both sides is because the original equation is x² - 30x = - 125, and so we must make sure the left hand side is still equal to x² - 30x.

So now we know that (x - 15)² multiplies out to give x² - 30x +225, we can write this as (x - 15)² in our equation:

(x² - 30x + 225) - 225 = - 125

is the same as:

(x - 15)² - 225 = - 125

Now add 225 to both sides of the equation:

(x - 15)² = - 125 + 225 = 100

(x -15)² = 100

The next step is to square root both sides. Be careful here, and remember that √100 can either be 10 or -10, as (-10)² = 100. To indicate both results, write ±10 ("plus or minus 10").

√(x - 15)² = √100

x - 15 = ±10

Because, there are two possible values for the right-hand side of the equation, we need to separate our equation into two equations:

1. x - 15 = 10

and

2. x - 15 = -10

Now we solve these two simple linear equations for x:

1. x = 10 + 15 <- add 15 to both sides

x = 25 This is our first solution.

2. x = -10 + 15 <- add 15 to both sides again

x = 5 This is our other solution.

So our two solutions are x = 5 and x = 25!