Pls answer both help asap

al receives $25 weekly allowance. he also babysits his two younger siblings on weekends and is paid $9 per hour. al's friend, Ma

Harlamova29_29 [7]

To answer this question you would set this up as an equation with all the information about Al on one side being equal to all the information for Matthew on the other side. This is what it would look like: $25 + $9h= $14h. h represents the number of hours they work. It they each worked 5 hours a week, they would make the same amount of money. I determined this by solving the equation for h. I subtracted 9h on both sides leaving you with $25=5h. H would equal 5 hours.

7 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 the percent increase from 3/8 to 7/8? HELP PLEASE I'VE BEEN STUCK ON THIS FOREVERRRRRRRR

iogann1982 [59]

3/8th = 0.375
<span>7/8th = 0.875
</span><span>The increase is over so, 1.000 would be .500 = 50%

Good luck with your studies. Hope this helps~!
</span>

3 0

3 years ago

Read 2 more answers

Select the correct answer from each drop-down menu.

vovikov84 [41]

Answer:

Its the same thing over and over again

B

Step-by-step explanation:

7 0

3 years ago

Solve for x/2=17 a.x=34 b.x=8.5 c.x=15 d.x=19

Sidana [21]

A. x=34

2 x 17 = 34
34/2 = 17

4 0

3 years ago