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blagie [28]
3 years ago
9

Tina practiced piano for 15 hours last month and 45 hours this month A. Use multiplication to write a statement comparing the ho

urs Tina practice during the two months B. Use addition to write a statement comparing the hours Tina practice during the two months
Mathematics
1 answer:
KonstantinChe [14]3 years ago
4 0

Answer: The answer is (a) 3 × x = y   and   (b) x + x + x = y.

Step-by-step explanation: Let 'x' and 'y' denote the number of hours that Tina practised last month and this month respectively.

Then, according to the given information, we have

x = 15   and   y = 45.

We can write the relation between 'x' and 'y' in two forms as follows -

(a) Multiplication form :   3 × 5 = 15  implies 3 × x = y.

(b) Addition form : 15 +15 +15 = 45  implies x + x + x = y.

Thus, the answers are (3 × x = y) and  (x + x + x = y).


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Find the slope of the line passing through the two points.<br><br> (2.1, 3.8), (3.1, 7.6)
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Answer:

3.8

Step-by-step explanation:

Slope formula- m = y2 - y1            

                                x2 - x1      

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3 years ago
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RoseWind [281]

Answer:

\displaystyle m=2

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Reading a Cartesian plane
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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Find points from graph.</em>

Point (-5, 0)

Point (-9, -8)

<u>Step 2: Find slope </u><em><u>m</u></em>

Simply plug in the 2 coordinates into the slope formula to find slope<em> m</em>

  1. Substitute in points [Slope Formula]:                                                            \displaystyle m=\frac{-8-0}{-9+5}
  2. [Fraction] Subtract/Add:                                                                                 \displaystyle m=\frac{-8}{-4}
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ra1l [238]
We need to simplify \frac{ \sqrt{14x^3} }{ \sqrt{18x} }

First lets factor \sqrt{14x^3}

\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3}
\sqrt{14} =  \sqrt{2} \sqrt{7} by applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a} \sqrt[n]{b}
\sqrt{x^3} = x^{3/2} By applying the radical rule \sqrt[n]{x^m} = x^{m/n}

So
\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3} = \sqrt{2} \sqrt{7}x^{3/2}

Now let's factor \sqrt{18x}
By applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a}  \sqrt[n]{b},
\sqrt{18x} =  \sqrt{18} \sqrt{x}
\sqrt{18} =  \sqrt{2} * 3

So \sqrt{18x} = \sqrt{2}*3 \sqrt{x}

So  \frac{ \sqrt{14x^3} }{ \sqrt{18x} } = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}  }

We know that \sqrt[n]{x} = x^{1/n} so \sqrt{x} = x^{1/2}

We now have \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}

We know that \frac{x^a}{x^b} = x^{a-b}
So \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x

We now got \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}&#10;

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} =   \frac{ \sqrt{7}x }{3}


All in All, we get \frac{ \sqrt{14x^3} }{ \sqrt{18x} } =  \frac{ \sqrt{7}x }{3}

Hope this helps! :D


6 0
3 years ago
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