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3 years ago

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Disadvantages of mathematical models are the model is a _____ of the real problem and does not include all _ of the prob

lemThe model may also only work in certain _______

1 answer:

Nonamiya [84]3 years ago

3 0

Answer:

Disadvantages of mathematicsl models are given.

Step-by-step explanation:

Disadvances of mathematicsl models are the model is a limited or bounded cases which taken from huge number of situations by ignoring many unknown facts of the real problem and does not include all important parameters or trivial parameters for accuricy of the result of system dynamics which gives a better result of the problem. The model may also only work in certain major state variables and parameters which may effect largly on the whole system.

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7(x+2)+3(x+5)=14(x+1)+2

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7x + 14 + 3x + 15 = 14x + 14 + 2 10x + 14 = 14x + 16 14 - 16 = 14x - 10x -2 = 4x X = - 2/4 X = - 1/2 OR X = - 0.5

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son4ous [18]

Answer:

The volume is 196 cubic yards

Step-by-step explanation:

Given

$Base\ Area = 7yd * 2yd$

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Step-by-step explanation:

Using the Pythagorean theorem,

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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 $a_i=0$ .

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