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DochEvi [55]
3 years ago
14

A chef uses 3 3/4 pounds of semolina flour and 1 5/8 pounds whole wheat flour for each batch of pasta she makes. One week she us

es a total of 86 pounds of flour.
Write an equation to solve for b, the number of batches of pasta the chef makes during the week.
How many batches of pasta does she make? Show your work.
Mathematics
1 answer:
atroni [7]3 years ago
5 0
One batch = 3 3/4 + 1 5/8
one batch = 3 6/8 + 1 5/8
one batch = 4 11/8
one batch = 5 3/8 pound of flour

One week = 86 pound 
86 ÷  5 3/8  = 86 ÷   43/8
86 ÷  5 3/8  = 86 x   8/43
86 ÷  5 3/8  =16

She makes 16 batches of pasta

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Since this can’t be factored easily, you have to use synthetic division. This makes the answer x+4+(-1/x+5)

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3 years ago
How do u do #22? Plz show me how u did it
hram777 [196]
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4 0
3 years ago
hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
Solve x + 3y = 9<br> 3x – 3y = -13 (1 point)
zavuch27 [327]

Answer:

(-1, 10/3)

x = -1

y = 10/3

Step-by-step explanation:

To solve a system, one of the methods you can use is <u>elimination</u>. To use elimination, you need <u>a variable to have the same coefficient in BOTH equations</u>. Since both equations have "3y", with the same coefficient (3), we can use this method.

We want to eliminate a variable by cancelling it out. Since positive 3y PLUS negative 3y is 0, the variable in eliminated. ADD each of the terms in the equations together.

.        x + 3y = 9

<u>+    3x – 3y = -13</u>

.     4x – 0y = -4           3y - 3y = 0        

.             4x = -4            Divide both sides by 4 to isolate 'x'

.               x = -1              Solved for the x-coordinate

Since we know one variable, 'x', we can easily find the other, 'y'. Substitute 'x' with -1 in any of the equations. Then, isolate 'x'.

x + 3y = 9

(-1) + 3y = 9

-1 + 1 + 3y = 9 + 1      Add 1 to both sides to cancel out left side.

3y = 9 + 1          Add on the right side (9 + 1 = 10)

3y = 10              Divide both sides by 3 to isolate 'y'

y = 10/3             Solved for the y-coordinate

Put the 'x' and 'y' coordinates together in an ordered pair, which you write as (x, y).

The solution to the system is (-1, 10/3).

5 0
3 years ago
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