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melomori [17]
2 years ago
13

Which of the following does NOT represent y as a quadratic function of X?

Mathematics
1 answer:
alex41 [277]2 years ago
7 0

Answer:

  • Top right equation

Step-by-step explanation:

<u>All the equations include the product of binomial or monomial of x but the top right one:</u>

  • y - 5 = x + 2y

This is a linear equation

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24% of what number is 12?
vladimir1956 [14]
To get that, you have to multiply the both numbers, which is 24*12=288
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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.  

5 0
3 years ago
Write the point-slope form of the equation of the line passing through the points (-5, 6) and (0, 1).
andrey2020 [161]
Formula of the slope of a linear function:


m = (y₂ - y₁)/(x₂ - x₁)

m= (-1 - 6)/[0- (-5)]

m= -7/5 This is the slope requested
7 0
3 years ago
A taxicab charges $1.75 for the flat fee and $0.25 for each mile. write an inequality to determine how many miles eddie can trav
aev [14]

The correct option is a. $1.75 + $0.25x ≤ $15; x < 53 miles.

The inequality that represents given situation is $1.75 + $0.25x ≤ $15.

<h3>What is inequalities?</h3>

An inequality compares the two values to determine if one is less than, larger than, or simply simply equal to the other.

  • a ≠ b indicates that an is not equal to b.
  • a < b indicates that an is less than b.
  • The expression a > b indicates that an is greater than b. (those two are called as strict inequality)
  • a ≤ b signifies that an is less than or equal to b.
  • The expression a ≥ b denotes that an is greater than or equal to b.

Now for the given question;

A cab charges $1.75 for the flat fee.

And, the cab charges $0.25 for each mile.

Total amount spent by Eddie is 1$15.

So, first and foremost. Because you have to pay the fixed charge and then pay for x miles is $0.25.

Thus, (15-1.75) / 0.25 = 53.

Therefore, the inequalities which describes the given scenario is;

$1.75 + $0.25x ≤ $15

where, x < 53 miles.

To know more about the inequalities, here

brainly.com/question/11613554

#SPJ4

The correct question is-

A cab charges $1.75 for the flat fee and $0.25 for each mile. Write and solve an inequality to determine how many miles Eddie can travel if he has $15 to spend.

a. $1.75 + $0.25x ≤ $15; x ≤ 53 miles

b. $1.75 + $0.25x ≥ $15; x ≥ 53 miles

c. $0.25 + $1.75x ≤ $15; x ≤ 8 miles

d. $0.25 + $1.75x ≥ $15; x ≥ 8 miles

3 0
1 year ago
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Solve 3x2+7x+1=0 using the quadratic formula.
NNADVOKAT [17]

Answer:

x=\frac{-7+-\sqrt{37} }{6}

Step-by-step explanation:

This (±) goes between the -7 and \sqrt{37}

3x^{2} +7x+1=0

ax^{2} +bx+c

-b ± \sqrt{b^2-4ac}/2a

-7 ± \sqrt{(7)^2-4(3)(1)}/2(3)

-7 ± \sqrt{49-12}/6

-7 ± \sqrt{37}/6

7 0
3 years ago
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