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3 years ago

Choose the equations that are not linear equations. - x - y = 2, x^2 + y = 4, x/2 - y^3 = 1, x = 5

Mathematics

1 answer:

tekilochka [14]3 years ago

7 0

Answer:

x^2 + y = 4
x/2 - y^3 = 1

Step-by-step explanation:

Any equation with a product of variables, or with a variable in an exponent or denominator, is <em>not a linear equation</em>. Here, "product of variables" includes powers and roots—any expression with an exponent other than 1, or a total of exponents other than 1.

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What is the common factor of the numerator and denominator in the expression <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%282x%2

mixer [17]

Answer:

Step-by-step explanation:

(x - 4)

What you are being asked to do is find the exact same binomial in the top as is in the bottom.

But there's a small catch. You must stipulate that x cannot equal 4. If it does, then you will get 0/0 which is undefined. You can't have that happening -- not at this level.

Any other value for x is fine.

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3 years 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 $a_i=0$ .

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3 years ago

Which of the following represents the zeros of f(x) = x3 − 11x2 + 38x − 40?

Delvig [45]

The answer is <span>5, 4, 2
</span>
Among all choices we have 5, so
x = 5
x - 5 = 0
Let's divide the expression by (x - 5) using the long division:
x³ - 11x² + 38x - 40
(x - 5) * x² = x³ - 5x² Subtract
____________________________
-6x² + 38x - 40
(x - 5) * (-6x) = -6x² + 30x Subtract
____________________________
8x - 40
(x - 5) * 8 = 8x - 40 Sutract
____________________________
0

Thus: x³ - 11x² + 38x - 40 = (x - 5)(x² - 6x + 8)

Now, let's simplify x² - 6x + 8.

x² - 6x + 8 = x² - 2x - 4x + 8 =
= x² - 2*x - (4*x - 4*2) =
= x(x - 2) - 4(x - 2) =
= (x - 4)(x - 2)

Hence:
x³ - 11x² + 38x - 40 = (x - 5)(x - 4)(x - 2)
To calculate zero:
x³ - 11x² + 38x - 40 = 0
(x - 5)(x - 4)(x - 2) = 0
x - 5 = 0 or x - 4 = 0 or x - 2 = 0
x = 5 or x = 4 or x = 2

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