What is hte vertex of f(x) = 5x^+20x-16

What is an exponent? Because i am confused

ollegr [7]

An exponent is a number written above and to the right side of a value or number in math. The value or number is called the base, and the expression is read out as;

"Base raised to the power of exponent."

An example is given below;

$10^3$

In this example, 10 is the base and 3 is the EXPONENT.

The expression is now read out as,

"10 raised to the power of 3"

Note that even unknown values (such x as we commonly use in math) can also be an exponent or base. For example, we can have;

$\begin{gathered} 10^x \\ OR \\ 2x^3 \end{gathered}$

So basically, an exponent indicates the number of times a number is used to multiply itself.

8 0

1 year ago

Miguel reels his teacher that 1/5 is the same as 20% what best justifies Miguel’s answer

horrorfan [7]

If you do the proportion, you have to put 1/5 = 20/100. The 20 represents the percentage. Then you have to say to yourself that how many time 5 goes to 100, well its x20 then if we multiply 1 x 20 we'll get 20. So, he is correct, 1/5 = 20%. Remember what number we think of when we hear the word percent, the number is 100.

5 0

3 years ago

Read 2 more answers

What is 2,621 divided by 37 with remainder

Dahasolnce [82]

Answer: 209

Step-by-step explanation:

3 0

2 years ago

I can’t find the the square centimeters of this

Temka [501]

Answer: 544?

Step-by-step explanation:

This is the answer if I'm supposed to solve the question in yellow

1/2*16(48+20) = 1/2*16*(68)

1/2*16=8

8*68 = 544

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