Use inverse operations to solve the equations -7×=119×=

Select each pair of functions that are inverses of each other

svlad2 [7]

Answer:

The first two choices only: A and B.

Step-by-step explanation:

The inverses are the ones with one relation having point (x,y) and the other relation having point (y,x). This most obvious when looking at points on a table or a list of points.

So the first pair are inverses because per point (x,y) on the first list you have (y,x) on the second list.

Examples:

1st list contains (-5,-9) while second list contains (-9,-5).

1st list contains (3,7) while the second list contains (7,3).

As long as (a,b) is in the first list and (b,a) is in the second or vice versa, then the pair of relations are inverses. So the first choice contains inverses.

Now lets talk about the pair of functions given in function notation.

Let's start with the first.

y=x+7

Extend the first idea more. Just swap x and y and then see after solving for y if what it equals is what g equals then f and g are inverses in the second choice.

x=y+7

Subtract 7 on both sides:

x-7=y

y=x-7

This is what g equals so the second choice is an answer.

The last choice does not contain a pair of inverses. Example: (2,3) is in the first list but (3,2) is not in the second.

6 0

3 years ago

Read 2 more answers

What types of numbers are undefined when they are under a radical sign? If you were dealing with the number √-1, would it be def

Misha Larkins [42]

The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1

If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1

If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)

If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)

however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1

If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1

and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)

Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.

8 0

2 years ago

The quotient of a number, divided by 3, equals-8

melisa1 [442]

Answer:

x / 3 = -8

Step-by-step explanation:

5 0

2 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

You start at 1, 4. You move left five units and down seven units. where do you end?

ololo11 [35]

-4,3 i think that’s the answer

6 0

2 years ago

Read 2 more answers