By factor decomposition and algebra properties we conclude that the fraction 42 / 49 is equivalent to the fraction 6 / 7.
<h3>Which pair of fractions is equivalent to each other?</h3>
According to the statement, we have the equation of a fraction and we need to find the expression by simplify the former expression. In this situation we must use factor decomposition and algebra properties to make simplification effective:
<h3>
</h3>
42 / 49 Given
(2 × 3 × 7) / (7 × 7) Factor decomposition
(2 × 3) / 7 × (7 / 7) Commutative and associative properties / Multiplication of fractions
6 / 7 × 1 Definition of division / Existence of the multiplicative inverse
6 / 7 Modulative property / Result
By factor decomposition and algebra properties we conclude that the fraction 42 / 49 is equivalent to the fraction 6 / 7.
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Answer:
4 1/3 cups
Step-by-step explanation:
All you have to do for this one is subtract 5 2/3 from 10, which is 4 1/3. Your equation could be 10-5 2/3=x.
Answer:
15
Step-by-step explanation:
A line of best fit is a line drawn at average distance to a series of scattered points. Thus, the correct option is:
Option B. The <em>data</em> point is 4. 5 units <u>below</u> the line of best fit.
A graph is said to be formed by plotted points on a <u>Cartesian </u>coordinate plane. When the plotted <em>points</em> are <u>scattered</u>, a <em>line of best fit</em> is required to shoe the relationship among the plotted <em>points</em>.
A <u>residual</u> <u>value</u> is a form of error that is determined from the <em>line of best fit</em>. A <u>line of best fit</u> is a line drawn at <em>average</em> point to a <em>series</em> of plotted <em>scattered</em> points.
Therefore, a residual value of 4.5 with respect to a <em>line of best fit</em> implies that: The data point is 4. 5 units <u>below</u> the <em>line of best fit</em>.
Thus the <em>correct</em> option is B.
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Step-by-step explanation:
<em>We</em><em> </em><em>can </em><em>see </em><em>that </em><em>they </em><em>are </em><em>vertically </em><em>opposite </em><em>angles </em><em>so </em><em>they </em><em>are </em><em>equal</em><em>. </em><em> </em><em>then</em>
<em>6a </em><em>+</em><em> </em><em>1</em><em>0</em><em>°</em><em> </em><em>=</em><em> </em><em>3a </em><em>+</em><em> </em><em>5</em><em>5</em><em>°</em>
<em>6a </em><em>-</em><em> </em><em>3a </em><em>=</em><em> </em><em>5</em><em>5</em><em>°</em><em> </em><em>-</em><em> </em><em>1</em><em>0</em><em>°</em>
<em>3a </em><em>=</em><em> </em><em>4</em><em>5</em><em>°</em>
<em>Therefore </em><em>a </em><em>=</em><em> </em><em>1</em><em>5</em><em>°</em>