M || n, find the value of x.<br> m<br> n<br> 65°<br> (4x-17)°<br> WILL MARK BRAINLIEST IF RIGHT!

If the following is an exponential growth function, f(x)=3(A)^x , what can the value of A be? Select all that apply.

mixer [17]

The answer can either be 1.25 or 2. As long as the value of A is greater than 1, the function is an exponential growth function.

6 0

3 years ago

Read 2 more answers

RR3-03 A posted speed limit of 55 mph means (Select the one BEST answer from the choices provided.) 8-2/43

Illusion [34]

A posted speed limit of 55 mph means that: You may drive 55 mph only under favorable conditions.

<h3>What is the speed limit?</h3>

Speed limits on road traffic, as used basically to set the legal maximum speed at which vehicles are permitted to utilize on a given stretch of road.

Now generally, speed limits are normally indicated on traffic signs reflecting the maximum allowable speed in kilometers per hour (km/h) or eevn miles per hour (mph). They are primarily set by the legislative bodies of national or provincial governments and then enforced by law enforcement agencies. This suggests that speed limits varies from road to road and is even absent in some certain areas.

Now, in this question we are told that the posted speed limit is 55 mph. Thus, from our definitions of speed limits above it is clear that A posted speed limit of 55 mph means that: You may drive 55 mph only under favorable conditions.

Read more about Speed Limit at; brainly.com/question/6457873

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4 0

2 years ago

6. 21 and 22 form a linear pair. If mZ1 = (5x + 9)* and m2 = (3x + 11)", find the measure of

VLD [36.1K]

Answer:

∠ 1 = 109° , ∠ 2 = 71°

Step-by-step explanation:

The 2 angles form a linear pair and sum to 180°

Sum the 2 angles and equate to 180

5x + 9 + 3x + 11 = 180 , that is

8x + 20 = 180 ( subtract 20 from both sides )

8x = 160 ( divide both sides by 8 )

x = 20

Then

∠ 1 = 5x + 9 = 5(20) + 9 = 100 + 9 = 109°

∠ 2 = 3x + 11 = 3(20) + 11 = 60 + 11 = 71°

3 0

3 years ago

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

8 0

3 years ago

What is 4 divided by 3/4?

musickatia [10]

Answer:

16/3 or 5.33

Step-by-step explanation:

Cross multiply

4/1 * 3/4

4 * 4 = 16

3 * 1 = 3

6 0

3 years ago

M || n, find the value of x. m n 65° (4x-17)° WILL MARK BRAINLIEST IF RIGHT!