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Setler79 [48]
3 years ago
11

Use cramer's rule to determine if the system is inconsistent system or contains dependent equations. 2x + 7 = 8 6x + 3y = 24 que

stion 24 options: system is inconsistent system contains dependent equations
Mathematics
2 answers:
Natali [406]3 years ago
8 0
Answer: In the equations that you have given, we have a dependent system.

2x + y = 8 (I assumed that you meant to type y instead of 7)
6x + 3y = 24


To use Cramer's Rule, we have to take the determinant of 3 different matrices written in the problem. Taking the determinant of the coefficient matrix produces a zero.

2  1   This is the coefficient matrix.
6  3

6 - 6 = 0

Since this is 0, the rest of the work will be undefined meaning the systems are dependent (or they are the versions of the same equation).
geniusboy [140]3 years ago
5 0

Answer:

The given system of equation is neither inconsistent nor contains the dependent equation.

Step-by-step explanation:

Given system of equation,

2x + 7y = 8,

6x + 3y = 24,

Since, the determinant of coefficients of variables,

D=\begin{vmatrix}2 & 7 \\ 6 & 3\end{vmatrix}

=2\times 3-7\times 6

=6-42

=-36

Since, D ≠ 0,

Hence, the given system of equation is neither inconsistent nor contains the dependent equation.

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Nady [450]
The answer is C

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7 0
3 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.
lakkis [162]

Answer:

2*sin(x)+y*cos(x)-cos(y)=C_1

Step-by-step explanation:

Let:

P(x,y)=2*cos(x)-y*sin(x)

Q(x,y)=cos(x)+sin(y)

This is an exact differential equation because:

\frac{\partial P(x,y)}{\partial y} =-sin(x)

\frac{\partial Q(x,y)}{\partial x}=-sin(x)

With this in mind let's define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x}=P(x,y)

and

\frac{\partial f(x,y)}{\partial y}=Q(x,y)

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y)

f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}

Now, let's replace the previous result into \frac{\partial f(x,y)}{\partial y}=Q(x,y) :

cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)

Solving for \frac{dg(y)}{dy}

\frac{dg(y)}{dy}=sin(y)

Integrating both sides with respect to y:

g(y)=\int\ sin(y)  \, dy =-cos(y)

Replacing this result into f(x,y)

f(x,y)=2*sin(x)+y*cos(x)-cos(y)

Finally the solution is f(x,y)=C1 :

2*sin(x)+y*cos(x)-cos(y)=C_1

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