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snow_lady [41]
2 years ago
5

Mr.Jamison brought 15 tubs of paint for the art classes he teaches for his Elementary kids. He paid $63.90 for the paint.If each

tube of paint costs the same amount, what was the price of one tube?​
Mathematics
1 answer:
xeze [42]2 years ago
6 0

Answer:

$4.26

Step-by-step explanation:

63.90 divide by 15 = $4.26

the price of one tube is $4.26

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If the price for 12 cookies is 42 dollars how much would if cost for 1 cookie
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$3.50

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42 divided by 12 is the price of one cookie.

42 divided by 12 = $3.50

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A computer is normally $800 but is discounted to $525. What percent of the original price does Mark pay? Round to a
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60 POINTS!!!! PLEASE HELP URGENT
sammy [17]

Using linear combination method to solve the system of equations 3x - 8y = 7  and x + 2y = -7 is (x, y) = (-3, -2)

<h3><u>Solution:</u></h3>

Given that, a system of equations are:

3x – 8y = 7 ⇒ (1) and x + 2y = - 7 ⇒ (2)

We have to solve the system of equations using linear combination method and find their solution.

Linear combination is the process of adding two algebraic equations so that one of the variables is eliminated. Addition or subtraction can be used to perform a linear combination.

Now, let us multiply equation (2) with 4 so that y coefficients will be equal numerically.

4x + 8y = -28 ⇒ (3)

Now, add (1) and (3)  

3x – 8y = 7  

4x + 8y = - 28  

----------------

7x + 0 = - 21  

7x = -21  

x = - 3  

Now, substitute "x" value in (2)

(2) ⇒ -3 + 2y = - 7  

2y = 3 – 7  

2y = - 4  

y = -2

Hence, the solution for the given two system of equations is (-3, -2)

3 0
3 years ago
How many Solutions does this system have? (1 point)
mixas84 [53]

The given system of equation that is 2x+y=3 and 6x=9-3y has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}

Let us first bring the equation in standard form for comparison

\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

To check how many solutions are there for system of equations a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0, we need to compare ratios of \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}

In our case,  

a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3

a_{2}  = 6, b_{2} = 3,\text{ and }c_{2} = -9

\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}

As \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

5 0
3 years ago
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