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Elodia [21]
2 years ago
8

Jesse's dog Angel weighs 18 pounds. How many ounces does angel weigh?

Mathematics
2 answers:
eimsori [14]2 years ago
8 0

Answer:

288 oz

Step-by-step explanation:

sineoko [7]2 years ago
5 0

Answer:288

Step-by-step explanation:

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V(t)=875-50t
Mandarinka [93]
Use socratic instead of this
6 0
3 years ago
Suppose you have 15 muffins. How many muffins are left after you give a friend 1/3 of them?
madam [21]
You would have 10 muffins left because 1/3 of 15 is 5 so 15-5=10
5 0
3 years ago
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Please help with Letter C
Papessa [141]
Your reasoning is correct, but you made a mistake with your long division.

1248/98 = 12 remainder 72

       __12_
98 |  1248
        -  98
        --------
           268
          -196
         --------
              72


Final Answer:  13 is minimum number of showings
6 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
What is the slope of a line that is parallel to the line y =3/4 x + 2
lbvjy [14]
Y=mx+b
m=slope

slopes of parallel lines are the same


y=3/4x+2
the slope is 3/4

the slope of a line parallel to that is 2/3
7 0
3 years ago
Read 2 more answers
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