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lesya692 [45]
3 years ago
11

Which ordered pair is in the solution set of the system of linear inequalities graphed below?

Mathematics
1 answer:
motikmotik3 years ago
5 0

Answer:

Graphed?

Step-by-step explanation:

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MATHS. SOMEONE, PLEASE HELP I AM PRETY DESPERATE.
valentina_108 [34]

Answer:

60

Step-by-step explanation:

the lowest where they all = the same is 60

7 0
3 years ago
Taylor tried to solve an equation step by step.
polet [3.4K]
B: step 2
You need to multiply the 7 with the 4.5
3 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
Solve for W -5.1=1.3+w/4
Lera25 [3.4K]
-5.1=1.3+w/4
Subtract 1.3 from both sides
-6.4=w/4
Multiply both sides by 4
-25.6=w

Check work:
-5.1=1.3+(-25.6/4)
-5.1=1.3+(-6.4)
-5.1=1.3-6.4
-5.1=-5.1

Hope this helps!  :)
7 0
3 years ago
Round 24,061,562 to the nearest 1,000 and 100,000
jenyasd209 [6]

Nearest 1,000 = 24,062,000

Nearest 100,000= 24,100,000

Hope its correct :)

6 0
3 years ago
Read 2 more answers
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