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olchik [2.2K]
3 years ago
6

What are the domain and range of the exponential function below F(x)=4^x+3

Mathematics
1 answer:
topjm [15]3 years ago
5 0
Given that the function f(x)=4^x+3 is an exponential growth question, it implies that the graph of the function is increase from negative infinity to positive infinity. Therefore we conclude that:
the domain is :
All real numbers.
The range will be all real number greater than 3. 
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Factoring trinomials
Kobotan [32]

Answer:

Trinomial

Step-by-step explanation:

Hope it helps

3 0
3 years ago
What is the inverse of function F f(x) 3-x/7
algol [13]

Answer: f^-1(x)=-7x+3 maybe ? i think

Step-by-step explanation:

8 0
3 years ago
Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t
schepotkina [342]

Answer:

(A) Set A is linearly independent and spans R^3. Set is a basis for R^3.

Step-by-Step Explanation

<u>Definition (Linear Independence)</u>

A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.

<u>Definition (Span of a Set of Vectors)</u>

The Span of a set of vectors is the set of all linear combinations of the vectors.

<u>Definition (A Basis of a Subspace).</u>

A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.

Given the set of vectors  A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right) , we are to decide which of the given statements is true:

In Matrix A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right) , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. R^3 has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans R^3.

Therefore Set A is linearly independent and spans R^3. Thus it is basis for R^3.

8 0
3 years ago
Find the vertex of the graph of the function.
lapo4ka [179]

Answer: The correct options are  (1) (5,10), (2) (3,-3), (3) x = -1, (4) y=(x+2)^2+3, (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

f(x)=a(x-h)^2+k        .....(1)

Where,  (h,k) is the vertex of the parabola.

(1)

The given equation is,

f(x)=(x-5)^2+10

Comparing this equation with equation (1),we get,

h=5 and k=10

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

f(x)=3x^2-18x+24

f(x)=3(x^2-6x)+24

To make perfect square add (\frac{b}{2a})^2, i.e., 9. Since there is factor 3 outside the parentheses, so subtract three times of 9.

f(x)=3(x^2-6x+9)+24-3\times 9

f(x)=3(x-3)^2-3

Comparing this equation with equation (1),we get,

h=3 and k=-3

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

f(x)=4x^2+8x+7

f(x)=4(x^2+2x)+7

To make perfect square add (\frac{b}{2a})^2, i.e., 1. Since there is factor 4 outside the parentheses, so subtract three times of 1.

f(x)=4(x^2+2x+1)+7-4

f(x)=4(x+1)^2+3

Comparing this equation with equation (1),we get,

h=-1 and k=3

The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.

(4)

The given equation is,

y=x^2+4x+7

To make perfect square add (\frac{b}{2a})^2, i.e., 2^2.

f(x)=x^2+4x+4+7-4

f(x)=x^2+4x+4+7-4

f(x)=(x+2)^2+3

Therefore, the correct option is  4.

(5)

The given equation is,

h=-16t^2+672t

It can be written as,

h=-16(t^2-42t)

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

x=\frac{b}{2a}

x=\frac{42}{2}

x=21

Therefore,  after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

f(x)=3x^3-12x^2-15x

f(x)=3x(x^2-4x-5)

Use factoring method to find the factors of the equation.

f(x)=3x(x^2-5x+x-5)

f(x)=3x(x(x-5)+1(x-5))

f(x)=3x(x-5)(x+1)

Equate each factor equal to 0.

x=0,-1,5

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.

3 0
3 years ago
99 POINT QUESTION PLUS BRAINLIEST!!!
MA_775_DIABLO [31]
The graphs arey=(1/3)x^2
and y=3
they intersect at x=3
so the limits are x=0 and 3
to rotate it about y=k, minus k from every function


area=\pi  \int\limits^3_0 {( \frac{1}{3}x^2-3)^2-(3-3)^2 } \, dx
which evaluates to 72/5pi

A is asnwer
7 0
3 years ago
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