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N76 [4]
3 years ago
11

What is the product 1/4*3 2/3

Mathematics
2 answers:
Scrat [10]3 years ago
6 0

Answer:

<h2>0.5</h2>

Step-by-step explanation:

Diano4ka-milaya [45]3 years ago
3 0

Answer:

11/12 is the product of 1/4*3 2/3

Step-by-step explanation:

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to make sure you understand the question

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1. Which of the following is NOT true about an isosceles trapezoid?
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On an isosceles trapezoid, the two sides that are not parallel to each other will be exactly the same length.  If this is true, than it would be create a symetric trapezoid.  The diagonals would be the same length.  The bases of any trapezoid are parallel, so this is true.  The diagonals cannot possibly be perpendicular because the 2 nonparallel sides would be slanted.  So, the answer is the 3rd choice.
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3x+1=7-4y solve for y pls help lol
Westkost [7]

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y = -3x/4+3/2

Step-by-step explanation:

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2 years ago
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Ntate runs 12 km in 3 hours. how many hours will it take him to run in 1 km​
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15 minutes or 1/4 hour

Step-by-step explanation:

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5 0
2 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
2 years ago
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