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Savatey [412]
3 years ago
11

6. At the beginning of school, Julie bought one 3-subject notebook and four single-subject notebooks

Mathematics
1 answer:
larisa [96]3 years ago
3 0

Answer: 3-subject notebook cost $0.67 or 67 cents and the single-notebook cost  $0.33 or 33 cents.

Step-by-step explanation:

Let's represent the 3-subject notebook by y and the single-notebook by x.

y has to cost twice as much as x and we can represent that by the equation

y = 2x  

We now know that the total cost of the 3-subject notebook and the single-subject notebook has to equal  $1.

We can also represent that by the equation and solve for x

2x + x = 1  

3x = 1

x = 0.33   $0.33 or 33 cents  

If the single-subject is $33 cents then subtract it from a dollar to see how much the 3-subject notebook costed.

1- 0.33 = 0.67

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4 0
3 years ago
In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv
ycow [4]

Answer:

• David

,

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

,

• David's House, B(15,13)

,

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Owen's distance from school (AC)

\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}

David's distance from school (BC)

\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

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1 year ago
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Step-by-step explanation:

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<span>1 and 2/3 time 1 and 1/3 is 2.22222222222 which rounds to 2.</span>
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