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Ivenika [448]
3 years ago
5

If the domain is {0, 2, -6}, what is the range of y = -2x + 3?

Mathematics
2 answers:
jasenka [17]3 years ago
6 0

Answer:

what is a domain

Step-by-step explanation:

please explain sorry

schepotkina [342]3 years ago
5 0

Answer:

the rage is 1 this should be the answer

Step-by-step explanation:

Y=-2x+3?

Y= -2 -2

X=1

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Which of these is a correct statement?
Montano1993 [528]

Answer:

the answer is B because it has only one unknown and also it is a linear equation so it will have only one answer which is (2 )

6 0
3 years ago
Simplify without using the absolute value sign<br><br> | x- ( - 18) | if x &gt; - 18
slega [8]
The expression would be > 0.

Subtracting a negative is the same as adding a positive; therefore -18--18 = -18+18 = 0.  Since we know that x>-18, 0 is below anything we could get from this.
7 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
A company has 200 employees. The company’s owner randomly chose 30 employees for an in-depth survey. Of those surveyed, 12 emplo
andrey2020 [161]
(this shows the percent of people that would work over time on a regular basis out of the 30 people) 12/30= (40%) of the 30 said they would work overtime.

200×.4=80 employees of the 200 would have probably have said they would work overtime.
8 0
3 years ago
-8, 22, 52, 82, ...<br> Find a 20
gladu [14]

Answer:

Step-by-step explanation:

It is an arithmetic sequence with common difference d = 30.

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8 0
3 years ago
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