All categories

3 years ago

The doctor recommends rest if the patient has the flu. The doctor recommends rest.

1 answer:

7 0

Illnesses ▬▬▬> Flu Other Illnesses Other Illnesses
\ (requiring (not requiring
\ rest) rest)
\ / |
\ / |
\ / |
Remedy ▬▬▬▬▬▬> Rest Other Remedies

- If the doctor recommends rest, the patient has the flu. ... FALSE

- The patient has the flu. ... (same thing) ... FALSE

- The patient does not have the flu. ... FALSE

No conclusions can be drawn.

You might be interested in

2. R(1, 1), 5(1,7), Т(5,7), U(5, 1); k = 45 у о Help

Gekata [30.6K]

(5,7) because when you divide that’s the answer you get

8 0

3 years ago

EASY PLS HELP WILL GIVE BRAINLY If -11 + N = -11, then N is the _____.

Tems11 [23]

Answer:

SteIf -11 + N = -11 then N is the additive inverse because if we do process of elimination (POE) we can cross out C and D because this is a addition problem not a multiplication, and we know its not B because that's like 99 + 0 = 99 set, but instead its A because its just like -8 + 8 = 0. (If this was not a multiple choice question then tell me)

3 0

3 years ago

Read 2 more answers

The line integral of xydx×x^2y^3dy where c is the triangle with points (0,0) (1,0) (1,2) using green's theorem

lyudmila [28]

If $\mathcal C$ is the boundary of the triangle $D$ , then by Green's theorem

$\displaystyle\int_{\mathcal C}xy\,\mathrm dx+x^2y^3\,\mathrm dy=\iint_D\left(\frac{\partial(x^2y^3)}{\partial x}-\frac{\partial(xy)}{\partial y}\right)\,\mathrm dA$
$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2x}(2xy^3-x)\,\mathrm dy\,\mathrm dx$
$=\displaystyle\int_{x=0}^{x=1}x\int_{y=0}^{y=2x}(2y^3-1)\,\mathrm dy\,\mathrm dx$
$=\displaystyle\int_{x=0}^{x=1}x(8x^4-2x)\,\mathrm dx=\frac23$

4 0

3 years ago

The shortest side of a triangle is 9 greater than twice the shortest side of a similar triangle. If the ratio of similitude is 7

chubhunter [2.5K]

63:27 is the answer...

7 0

4 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