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3 years ago

What is the domain of the function?

Mathematics

2 answers:

STALIN [3.7K]3 years ago

8 0

The domain is the variation of values that x can equal. In this case, the domain is from 5 to infinity.

Sav [38]3 years ago

8 0

Answer:

Domain of function is [5,∞) or {x : 5 ≤ x < ∞ }.

Step-by-step explanation:

We are given with a Graph of a function.

We need to find Domain of the function in the graph.

We know that Domain of a Function is a set of elements of numbers for which we get a image from co-domain of the function or for which function is defines/ exist.

From graph Domain is the value of x-axis for which function is defined.

function is taking vale from x = 5 to x = ∞

Therefore, Domain of function is [5,∞) or {x : 5 ≤ x < ∞ }.

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The common difference of the arithmetic sequence

oksian1 [2.3K]

Answer:

its 1/3

Step-by-step explanation:

please mark it as brainliest!!<3

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2 years ago

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Pls help(ill mark brainlest)

gavmur [86]

Answer:

i think c

Step-by-step explanation:

7 0

2 years ago

Solve.<br><br> Round to the nearest hundredth.<br><br> 29.3 = 1.7y

Elena-2011 [213]

Steps to Solve:

29.3 = 1.7y

29.3/1.7 = 1.7y/1.7

y = 17.2352941176

In this case, we round up the hundredths place because the number in the thousandths place is 5 or higher.

y = 17.24

Hope this helps!! :)

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3 years ago

Read 2 more answers

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

Hey guys help me with this question plzzzzzz

Ad libitum [116K]

Answer:

625 cm³ and 64000 cm³

Step-by-step explanation:

Given

x = $\sqrt[3]{V}$ ( cube both sides )

x³ = V

(a)

x = 25 cm , then

V = 25³ = 625 cm³

(b)

x = 40 cm , then

V = 40³ = 64 000 cm³

5 0

2 years ago