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3 years ago

What two numbers multiply to 20 and add up to 17

1 answer:

4 0

That is not possible, since 20 has the factors of 1,2,4,5,10, and 20. None of those add up to 17, maybe your question has a mistake in it. For further questions, you can message me. Hope this helps! Please rate brainiest answer.

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Please help me answer it

jeka94

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Subtract these polynomials (3x^2x+4)-(x^2+2x+1)

harina [27]

2x^2+3. the 2x cancel out and the rest is combining like terms

6 0

3 years ago

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If p is a true statement and q is false, what is the truth value of p ∨ q?

ZanzabumX [31]

P V q = T V F = T

so, pVq is true.

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

Determine which equation is parallel to the line JK and which is perpendicular to like JK

sattari [20]

Answer:

If the equation of JK is y = mx + c, then y = mx + c' is the equation of the straight line parallel to above and $y = - \frac{1}{m}x + b$ will be perpendicular to that.

Step-by-step explanation:

Let us assume that the line JK has the equation in slope-intercept form as

y = mx + c ............. (1)

Therefore, the equation has slope = m

Now, any straight line having an equation with slope m will be parallel to line JK.

So, the equation of the straight line which is parallel to equation (1) will be

y = mx + c', where, c' is any real constant.

Now, let us assume that another straight line having equation

y = nx + a is perpendicular to the line JK i.e. equation (1).

Now, we know if two lines are perpendicular to each other then the product of their slopes will be - 1.

So, mn = - 1

⇒ $n = - \frac{1}{m}$

Therefore, the equation of a straight line which is perpendicular to equation (1) will be $y = - \frac{1}{m}x + b$ where b is any real constant. (Answer)

3 0

3 years ago

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