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3 years ago

WILL GIVE BRAINLIEST!!!

Mathematics

2 answers:

Olin [163]3 years ago

5 0

Answer:

c would be your answer

Step-by-step explanation:

Paladinen [302]3 years ago

3 0

It would be c love! xx

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Simplify 36*3. O A. 35 B. 36 O C. 37 O D. 96 HELP ME

Lelu [443]

Answer:

12/1

Step-by-step explanation:

hope this helped if not sorry

6 0

3 years ago

Kyle is stringing a necklace with beads he puts black beads on 5/8 of the string and white beads on 1/4 of the string Kyle think

Alexeev081 [22]

Answer:Kyle’s claim is not reasonable

Step-by-step explanation:

Total length of the string of the necklace that Kyle covered with black beads is 5/8.

Total length of the string of the necklace that Kyle covered with white beads is 1/4.

Total length of the string of the necklace that Kyle covered with black beads and white beads is would be

1/4 + 5/8 = 7/8 = 0.875

Kyle thought that he will cover 6/12 of the string with beads. 6/12 = 0.5

It means that he covered more than 6/12. Kyle's claim was wrong

3 0

3 years ago

V/199 is between...?

MissTica

I think E but I’m probably wrong???

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

In which set do all the values make the inequality 2x-1<10 true?

GenaCL600 [577]

Answer:

The right answer is letter D. (2,3,4)

5 0

3 years ago

Read 2 more answers