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marshall27 [118]
2 years ago
7

What is 1 cup equal to

Mathematics
2 answers:
dezoksy [38]2 years ago
8 0
A lot of things what thing to you need it converted to?
9966 [12]2 years ago
3 0
8 ounces. 123123123123123123

You might be interested in
Jonny has five Apples he gives four to Sarah how many apples do he have left ?
Korolek [52]

Answer:

He has one.

Step-by-step explanation:

He has five, and gives sarah four. That means that you take all but one away from 5. 5-4=1.

6 0
3 years ago
The maximum height of a vehicle that can safely pass under a bridge is 12 feet 5 inches. A truck measures 162 inches in height.
Nutka1998 [239]
It would not pass under the bridge because ig you divide 162 by 12 it would equal 13.5
4 0
3 years ago
Create a linear equation in the slope-intercept form that contains points (2,−8) and (6,−4) .
nadezda [96]

Answer:

y = x - 10

Step-by-step explanation:

Find the slope:

-4 - (-8) ÷ 6 - 2

-4 + 8 ÷ 6 - 2

4 ÷ 4

= 1

Substitute into the slope-intercept equation:

y = mx + b

Let's use the first coordinate (2,-8) for the x and y values

m = 1

-8 = 1(2) + b

-8 = 2 + b

-2    -2

-10 = b

Write equation:

y = 1x -10 or y = x -10

3 0
3 years ago
Brainlest 5 starred and thanked daily if you get this correct<br> Try your hardest
mart [117]
The cycle number two is the answer
8 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
2 years ago
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