What's 23/4 on a number line?

Edmentum// monomials

stepladder [879]

Answer:

625b^6

Step-by-step explanation:

= 5^3 * 5b^6

= 5^3 * 5

= 625

6 0

3 years ago

A player is randomly dealt a sequence of 13 cards from a deck of 52-cards. All sequences of 13 cards are equally likely. In an e

VikaD [51]

Answer:

Answer is 1/13

Step-by-step explanation:

Here we are not intimated about the first (01) twelve (12) cards that are shown, the probability that the thirteenth (13) card dealt is a King, if this is the same as the probability that the (01) first card shown, or in fact any particular card dealt is a King, then it equals:

=4/52

=1/13 ANS

5 0

3 years ago

Order least to greatest 1/2, 0.55, 5/7

Nataliya [291]

1/2<0.55<5/7
This is the order from least to greatest

8 0

3 years ago

Read 2 more answers

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

The answer is 48 can someone explain how?

mestny [16]

Answer:

Step-by-step explanation:

(1)

apple × apple × 2apple = 54

2apple³ = 54 divide both sides by 2

apple³ = 27 → apple = ∛27

apple = 3

(2)

apple + apple × green_apple = 24 substituite apple = 3

3 + 3 × green_apple = 24 subtract 3 from both sides

3 × green_apple = 21 divide both sides by 3

green_apple = 7

(3)

strawberry - bannana - banana = 0 → strawberry = 2banana

(4)

strawberry + banana + cherry = 24 substitute from (3)

2banana + banana + cherry = 24

3banana + cherry = 24

(5)

apple + banana - cherry = -1 substitute apple = 3

3 + banana - cherry = -1 subtr3 from both sides

banana - cherry = -4

Add both sides of (4) and (5)

3banana + cherry = 24

+banana - cherry = -4

4banana = 20 divide both sides by 4

banana = 5

Substitute it to (4):

3(5) + cherry = 24

15 + cherry = 24 subtract 15 from both sides

cherry = 9

Substitute to the last equation:

3 + 5 × 9 = 3 + 45 = 48

/USED PEMDAS/

7 0

3 years ago