Jennifer has 15 plates and 18 napkins that she wants to arrange in baskets. There needs to be the same number

a variable is like a placeholder, an empty spot on the pans. and the expressions on both sides are kind of the conditions for the things to put in these empty spots so that the status of the scale stays unchanged.

for an equation the scale has to stay balanced all the time.

and in an inequality one side is heavier that the other (it normally does not matter how much heavier) with "<" or ">" signs, or the scale can be also balanced with "<=" or ">=" signs. just the other side must never be heavier.

A

so, in our case here

2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = 4. no other value of x allows the scale to be balanced.

2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = 4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. 5 ?

2×5 - 5 >= 3

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = 5 is also a valid solution.

what else ?

let's simplify the inequality

2x - 5 >= 3

2x >= 8

x >= 4

so, really every value of x that is greater or equal to 4 is a valid solution for the inequality.

B

-2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = -4. no other value of x allows the scale to be balanced.

-2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = -4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. -5 ?

-2×-5 - 5 >= 3

remember, -×- = +

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = -5 is also a valid solution.

what else ?

let's simplify the inequality

-2x - 5 >= 3

-2x >= 8

x <= -4

a multiplication or division by a negative value flips the inequality sign, because such an operation makes a light weight heavy and a heavy weight light.

so, really every value of x that is less or equal to -4 is a valid solution for the inequality.

3 0

2 years ago

Robert is writing a paper for language arts class he can write one third of the page in 1/4 of an hour at what rate is Robert wr

Mars2501 [29]

He's writing one paper every 3/4 of an hour

1/3: 15 minutes
2/3: 30 min
3/3: 45 minutes

Or, one paper every 45 minutes/ 3/4 of an hour

7 0

3 years ago

Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is $g^{-1}(x) = e^{x}$

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is $g^{-1}(x) = e^{x}$

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y = ln(x)

Taking exponents of both sides, we have

$e^{y} = e^{lnx} \\e^{y} = x$

Replacing x with y, we have

$y = e^{x}$

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is $g^{-1}(x) = e^{x}$

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0

2 years ago