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2 years ago

A linear function passes through (-2, 12) and (4,0). What is the slope of the function?

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

5 0

Answer:

-2

Step-by-step explanation:

I actually just did functions in math at the end of last semester I offer you my condolences dude

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Cual es la respuesta de 5762÷34

juin [17]

is is 169.47058823 yes this the anser

7 0

3 years ago

Which number line shows the solution set for StartAbsoluteValue 2 p minus 4 EndAbsoluteValue greater-than-or-equal-to 6?

maks197457 [2]

Answer:

can u add a picture?

Step-by-step explanation:

8 0

2 years ago

While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at

tekilochka [14]

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

<h3>How to estimate the speed of the moving walkway relative to the airport terminal?</h3>

Let x be the speed of the walkway.

(2.8 + x) = speed of child moving in direction of the walkway

(2.8 - x) = speed of child moving against the direction of the walkway

Travel time = distance/speed

Travel time of child moving in direction of walkway = 23/(2.8+x)

Total elapsed time given = 29s

23/(2.8 + x)+ 23 / (2.8-x) = 29

LCD = (2.8 + x)(2.8 - x)

$23(2.8 - x) + 23(2.8 + x) = 29(2.8 + x)(2.8 -x)$

simplifying the equation, we get

$23*2.8-23x+23*2.8+23x=29(2.8^2-x^2)$

$23(2.8+2.8)/29=2.8^2-x^2$

$x^2=(2.8)^2-(23*5.6)/29)=3.4$

$x=\sqrt{3.4}=1.84m/s$

Speed of walkway = 1.84 m/s

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

To learn more about Speed refer to:

brainly.com/question/4931057

#SPJ4

4 0

1 year ago

I need an answer quick. I am trying to finish this fast.

inessss [21]

Answer:

D, m<4 is 137°

Step-by-step explanation:

The answer would be D, because m<4 would be supplementary with the angle that's 43 degrees. Supplementary angles add up to 180, so this could be found through the following equation.

x + 43 = 180

x would represent <4

You would now subtract 43 from both sides.

x = 137

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