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3 years ago

9

Jamie has 7/10 of a pound of chicken to cook for dinner if she's going to split the chicken between three people how much chicke

n with each person get

2 answers:

DiKsa [7]3 years ago

8 0

we know that

Jamie has $7/10$ of a pound of chicken to cook for dinner for three people

so

<u>Step $1$ </u>

Divide $7/10$ by $3$

$\frac{(7/10)}{3} =\frac{7}{10*3} = \frac{7}{30}\ lb$

therefore

<u>the answer is</u>

Each person get $\frac{7}{30}\ lb$ of chicken

ss7ja [257]3 years ago

6 0

All you have to do is divide 7/10 and 3. the answer is 7/30 pounds

You might be interested in

There are two types of tickets sold at the Canadian Formula One Grand Prix race. The price of 4 grandstand tickets and 2 general

padilas [110]

Answer:
*a general admission ticket is $160
*a grandstand ticket is $390

Explanation: make 2 equations from the provided information
**g is a grandstand ticket and a is a general admission ticket.

4g+2a=1880, 4g+4a=2200
Steps: 4g+2a=1880 changes to 4g=1880-2a because you move the 2a in order to solve for g.
You can substitute the 4g into 4g+4a=2200 to get 1880-2a+4a=2200 and simplify to get 1880+2a=2200. subtract 1880 from both sides and you get 2a=320. divide both sides by 2 to get a=160.
After that you can substitute the value for a into 4g+4a=2200 to get 4g+4(160)=2200. simplify to get 4g+640=2200. subtract 640 from both sides to get 4g=1,560. divide both sides by 4 to get g=390.

Hope this helped! :)

7 0

3 years ago

Convert the given system of equations to matrix form

yuradex [85]

Answer:

The matrix form of the system of equations is $\left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right] \left[\begin{array}{c}x&y&w&z&u\end{array}\right] =\left[\begin{array}{c}5&4&3\end{array}\right]$

The reduced row echelon form is $\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

The vector form of the general solution for this system is $\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Step-by-step explanation:

<em>Convert the given system of equations to matrix form</em>

We have the following system of linear equations:

$x+y+w+z-3u=5\\x-y-2w+z+2u=4\\2x+w-z+u=3$

To arrange this system in matrix form (Ax = b), we need the coefficient matrix (A), the variable matrix (x), and the constant matrix (b).

so

$A= \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right]$

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]$

$b=\left[\begin{array}{c}5&4&3\end{array}\right]$

<em>Use row operations to put the augmented matrix in echelon form.</em>

An augmented matrix for a system of equations is the matrix obtained by appending the columns of b to the right of those of A.

So for our system the augmented matrix is:

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\1&-1&-2&1&2&4\\2&0&1&-1&1&3\end{array}\right]$

To transform the augmented matrix to reduced row echelon form we need to follow this row operations:

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\2&0&1&-1&1&3\end{array}\right]$

add -2 times the 1st row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\0&-2&-1&-3&7&-7\end{array}\right]$

multiply the 2nd row by -1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&-2&-1&-3&7&-7\end{array}\right]$

add 2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&2&-3&2&-6\end{array}\right]$

multiply the 3rd row by 1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&1&-3/2&1&-3\end{array}\right]$

add -3/2 times the 3rd row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 3rd row to the 1st row

$\left[\begin{array}{ccccc|c}1&1&0&5/2&-4&8\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 2nd row to the 1st row

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

<em>Find the solutions set and put in vector form.</em>

<u>Interpret the reduced row echelon form:</u>

The reduced row echelon form of the augmented matrix is

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

which corresponds to the system:

$x+1/4\cdot z=3\\y+9/4\cdot z-4u=5\\w-3/2\cdot z+u=-3$

We can solve for <em>z:</em>

<em> $z=\frac{2}{3}(u+w+3)$ </em>

and replace this value into the other two equations

<em> $x+1/4 \cdot (\frac{2}{3}(u+w+3))=3\\x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}$ </em>

$y+9/4 \cdot (\frac{2}{3}(u+w+3))-4u=5\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}$

No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The system has infinitely many solutions:

<em> $x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}\\z=\frac{2u}{3}+\frac{2w}{3}+2$ </em>

where <em>u</em> and <em>w</em> are free variables.

We put all 5 variables into a column vector, in order, x,y,w,z,u

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=\left[\begin{array}{c}-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}&\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}&w&\frac{2u}{3}+\frac{2w}{3}+2&u\end{array}\right]$

Next we break it up into 3 vectors, the one with all u's, the one with all w's and the one with all constants:

$\left[\begin{array}{c}-\frac{u}{6}&\frac{5u}{2}&0&\frac{2u}{3}&u\end{array}\right]+\left[\begin{array}{c}-\frac{w}{6}&-\frac{3w}{2}&w&\frac{2w}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Next we factor <em>u</em> out of the first vector and <em>w</em> out of the second:

$u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

The vector form of the general solution is

$\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

7 0

4 years ago

I will give you brainlest if you are the first one to answer(and the best)

Crank

Answer:

4 kg

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Ecuacion de parabola con foco (-2,5) directriz en x=3

Dmitry_Shevchenko [17]

The required equation of the parabola is (y-5)² = 20(x+2)

<h3>Equation of a parabola</h3>

The standard equation of a parabola is expressed as;

(y-k)² = 4p(x-h)

where;

vertex = (h, k)

focus = (p+k, k)

directrix x = -p + h

Note that the directrix given will determine the type of the equation of a parabola

Given the following parameters

focus = (-2, 5) = (h, k)

x = -p + h

3 = p -2

p = 5

Substitute

(y-k)² = 4p(x-h)

(y-5)² = 4(5)(x-(-2))

(y-5)² = 20(x+2)

Hence the required equation of the parabola is (y-5)² = 20(x+2)

Learn more on equation of parabola here; brainly.com/question/4061870

#SPJ1

6 0

2 years ago

What is the area of the object above

aev [14]

The area would be 90cm
Explanation: To find area u multiple length times width, therefor 12x7 = 84 and 2x3 = 6
Then add it all together and it will equal 90cm

6 0

3 years ago