All categories

1 year ago

What is the domain of the exponential function shown below?

Mathematics

1 answer:

Phoenix [80]1 year ago

7 0

The function will exist on all real numbers. Hence the domain of the exponential function given is all real numbers

<h3>Domain of a function</h3>

The domain are independent values of an expression for which it exist. The standard exponential equation is expressed as y = ab^x

Given the exponential function as shown:

y = 5(3)^x

The function will exist on all real numbers. Hence the domain of the exponential function given is all real numbers

Learn more on domain here: brainly.com/question/26098895

#SPJ1

You might be interested in

PLEASE HELP GUYS i am struggling so much, two questions

ankoles [38]

Answer: the equation of the standard parabola

1) $(y-6)^2 = 4 (x-1)$

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

Explanation

Parabola:-

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

The locus of P such that $\frac{SP}{PM} = constant$
The fixed point 'S' is called the Focus.
The fixed line'l 'is called the directrix of the conic
The constant ratio is known as the eccentricity, denoted by 'e'
If e=1 , the conic is called a parabola

1) Step 1 :-

Given the focus S = (2,6) and directrix is x=0

we know that $\frac{SP}{PM}=1$

now cross multiplication , we get

$SP = PM$

squaring on both sides,we get

$SP^{2} = PM^2$

step 2:-

now using distance formula is

$\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)$

Given S =(2,6) and P(x,y) be any point on parabola

$SP^2 = (x-2)^2+(y-6)^2$ ........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

$\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }$

Given the directrix is x =0 and P(x,y) be any point on parabola

$PM^2 = \frac{x^2}{\sqrt{1}^2 }$ ......(2)

equating equation(1) and equation (2), on simplification

we get $(x-2)^2+(y-6)^2 = x^2$ .....(3)

apply $(a-b)^2 = a^2+ b^2+2 ab$

now the equation (3) is

$(y-6)^2 = 4 x-4$

now the standard form of parabola is

$(y-k)^2 = 4 a(x-h)$

Final answer:-

$(y-6)^2 = 4 (x-1)$

2) Explanation:-

step 1:

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

$(x-h)^2 = 4 a (y-k)$

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of $AS = \sqrt{(-5-(-5)^2+(2-6)^2}$

on simplification we get a =4

Final answer :-

the vertex (h,k) = (-5,2) and a=4

$(x-h)^2 = 4 a (y-k)$

The standard parabola is $(x+5)^2 = 16 (y-2)$

5 0

3 years ago

Plzzzz help mee plzzzzzzz

yan [13]

Answer:

A. 20%

Step-by-step explanation:

Hope it helps!

8 0

2 years ago

Evaluate 3a²b + 1–20 if a=-5, b=9, and c=-10.

DIA [1.3K]

Answer:

=656

Step-by-step explanation:

Evaluate for a=5,b=9

3(52)(9)+1−20

ps. there is no c

6 0

3 years ago

Find the constant of proportionality k as a fraction in simplest form. then enter an equation for the relationship between x and

Temka [501]

Answer:

k=1/4 and y=(1/4)x

7 0

3 years ago

The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni

aleksandr82 [10.1K]

Answer:

a) $population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44$

b) $sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3$

$sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70$

c) both the means of the two samples underestimate the population mean.

the mean pulse rate of sample 1 underestimates the population mean.

the mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

i) population size, N = 9

a) $population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44$

b) $sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3$

$sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70$

c) both the means of the two samples underestimate the population mean.

the mean pulse rate of sample 1 underestimates the population mean.

the mean pulse rate of sample 2 underestimates the population mean.

3 0

3 years ago