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3 years ago

Tell whether the argument below is an example of inductive or deductive reasoning.

Mathematics

1 answer:

Lilit [14]3 years ago

6 0

Answer:

Deductive Reasoning

Step-by-step explanation:

We take the statement as true,

"...all dogs in the kennel are collies..."

we also take

- collies are a type of dog

and

- Rover is a dog

So if Rover is a dog, and all dogs in the kennel are collies, then Rover is a collie, because he is a dog and inside the kennel (after we take the statement all dogs in the kennel are collies).

This type of reasoning follows the Deductive Reasoning pattern, if A = B, and B is C then A is also C.

While inductive takes that if A = B, and C is B just like D is B, then E is also B.

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Please answer with proof please!!!

qwelly [4]

A
Cause it can’t be B cause she’s moving
And if it was D he would be slowing down and C doesn’t match up with the 8hours a day

3 0

2 years ago

Read 2 more answers

Simplify the radical expression. √45 + √20 - √5

viva [34]

√45 = √9√5 = 3√5
√20 = √4√5 = 2√5

3√5+2√5-√5=4√5

answer = 4√5
:)

5 0

3 years ago

Simply: 200+(20-4)-2×3

alexandr1967 [171]

Answer:

The answer is 210

Step-by-step explanation:

200+(20-4)-2x3

first you'd do 20-4= 16

then you'd do 2x3=6

next you'd 16-6=10

finally 200+10=210

5 0

2 years ago

Line segment AB has endpoints A(1,8) and B(7,−4). What are the coordinates of the point located 5/6 of the way from A to B?

const2013 [10]

The coordinate of the point is (6,-2)

<h3>How to determine the coordinate of the point?</h3>

The given parameters are:

A = (1,8)

B = (7,-4)

The location of the point (i.e 5/6) means that the ratio is:

m :n = 5 : (6 - 5)

m : n = 5 : 1

The coordinate of the point is then calculated as:

$(x,y) = \frac{1}{m + n}* (mx_2 + nx_1,my_2 + ny_1)$

So, we have:

$(x,y) = \frac{1}{5 + 1}* (5 * 7 + 1 * 1 , 5 * -4 + 1 * 8)$

Evaluate

$(x,y) = \frac{1}{6}* (36 , -12)$

Evaluate the product

(x,y) = (6,-2)

Hence, the coordinate of the point is (6,-2)

Read more abut line segment ratio at:

brainly.com/question/12959377

#SPJ1

6 0

1 year ago

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

3 years ago