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Lerok [7]
3 years ago
5

Draw rectangular fraction models of 3 x 1/4 and 1/3 x 1/4. Compare multiplying a number by 3 and by 1 third.

Mathematics
1 answer:
Andrew [12]3 years ago
3 0

Answer:

In the file are the rectangular fraction models.

In the second draw, you can observe that the result of multiply \frac{1}{3}\times\frac{1}{4} is obtained the following form:

1. Draw a rectangle and divide it into four pieces.

2. Then create the fraction by coloring one column of pink.

3. We split the rectangle in thirds by drawing horizontal lines that divide the rectangle into three equal rows.

4. Then we color one row with green to show one third of one quarter

5.  So the answer (product) is the area of the rectangle where both of the shadings overlap each other. The denominator corresponds to the total of squares and the numerator of the product is the number of small rectangles where the shadings overlap.

When we multiply a number by 3 the result is a number greater that the number that we multiply, but when we multiply by \frac{1}{3} the result is a number smaller that the number that we multiply.

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f(x)=40

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so if x=2 it would be 40/8 which is 5

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In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.​
Karo-lina-s [1.5K]

Answer:

BC=11\ cm

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

m\angle DBC=\frac{1}{2}[arc\ DC]

substitute the values

60\°=\frac{1}{2}[arc\ DC]

120\°=arc\ DC

arc\ DC=120\°

step 2

Find the measure of arc BC

we know that

arc\ DC+arc\ BC=180\° ----> because the diameter BD divide the circle into two equal parts

120\°+arc\ BC=180\°

arc\ BC=180\°-120\°=60\°

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

m\angle BDC=\frac{1}{2}[arc\ BC]

substitute the values

m\angle BDC=\frac{1}{2}[60\°]

m\angle BDC=30\°

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

sin(\angle BDC)=BC/BD

BC=(BD)sin(\angle BDC)

substitute the values

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6 0
3 years ago
Solve.
Nikitich [7]
The answer is C: 10\text{ }^1/_2. Here are the details:

\text{Equation:}\\ 6\text{ }^1/_3+10\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 6+10=16\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^1/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^2/_6+\text{ }^3/_6=\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Add 'em up!}\\
16\text{ }^5/_6

\text{Equation:}\\
16\text{ }^5/_6+3\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Start with the integers.}\\
16+3=19\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{...then with the fractions, but rewrite them first to make it easier.}\\
^5/_6+\text{ }^5/_6=\text{ }^{10}/_6\stackrel{\text{rewrite}}{\to}\text{ }^5/_3\stackrel{\text{rewrite}}{\to}1\text{ }^2/_3\stackrel{?}{=}26\text{ }^1/_6\\
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20\text{ }^2/_3

\text{Last equation:}\\ 20\text{ }^2/_3+5\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 20+5=25\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^2/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^4/_6+\text{ }^3/_6=\text{ }^7/_6\stackrel{\text{rewrite}}{\to}1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6\\
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25+1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6

26\text{ }^1/_6\stackrel{\checkmark}{=}26\text{ }^1/_6
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3 years ago
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Answer:

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Step-by-step explanation:

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