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3 years ago

Plz help

Mathematics

1 answer:

jasenka [17]3 years ago

3 0

Answer: the answer is 19.6

Step-by-step explanation:

just add up by each number, for example 5.6+3 miles is 8.6 miles so just keep adding up the miles to get this answer

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What is the equation of the line in point slope form that passes through the point (-1, -5) and has a slope of -2

daser333 [38]

Option 4.) y+1=-2(x+5)

7 0

3 years ago

I need help please i

balandron [24]

Answer:

2 7/24

Step-by-step explanation:

First we have to add the gallons of two cow of each owner:

Hank's cows: 4 3/4 + 4 1/8

Add the whole numbers:

4 + 4 = 8

Now for 3/4 and 1/8

Find common

6/8 + 1/8

= 7/8

Put together :

8 7/8

Debra's cows: 5 1/2 + 5 2/3

Add the whole number

5 + 5 = 10

Now for 1/2 and 2/3

3/6 + 4/6

1 1/6

10 + 1 1/6

= 11 1/6

Since now we know that:

Hank - 8 7/8

Debra - 11 1/6

we can solve the question:

How many more gallons of milk did Debra's two cows produce on that compared to Hank's two cows?

Solution:

11 1/6 - 8 7/8

Subtract whole number

11 - 8 = 3

Now for 1/6 and 7/8

4/24 - 21/24

4-21/24

-17/24

2 7/24

~lenvy~

7 0

2 years ago

Read 2 more answers

Helpp!!! very easy question

bogdanovich [222]

Answer:

The answer is D, E, F or it is also D to F

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Three pairs of gloves- a red pair, a blue pair, and a green pair-are in a drawer. If the gloves are removed at random without re

balandron [24]

Answer:

Step-by-step explanation:

Given that three pairs of gloves- a red pair, a blue pair, and a green pair-are in a drawer. The gloves are removed at random without returning any to the drawer.

We have to find the minimum number that must be removed in order to guarantee having a matched pair of gloves

No of different colours = 3 (red, blue, green)

Hence no of gloves that must be removed = 3+1

If 4 gloves are removed, only 3 can be of different colours 1 will have same colour as any one of the three.

So a pair of same colour would be obtained

Answer is 4

8 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