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Aleksandr [31]
3 years ago
10

Draw fraction rectangles on dot paper to solve the problems.

Mathematics
1 answer:
juin [17]3 years ago
6 0

Answer:

1) \frac{7}{20}

2) 1\frac{1}{12}

3) \frac{7}{8}

Step-by-step explanation:

1) Given problem,

\frac{3}{5}-\frac{1}{4}

LCM(5,4) = 20,

\frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}

\frac{1}{4}=\frac{1\times 5}{4\times 5}=\frac{5}{20}

\implies \frac{3}{5}-\frac{1}{4}=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}

2) Given problem,

\frac{5}{6}+\frac{1}{4}

LCM(6, 4) = 12,

\frac{5}{6}=\frac{5\times 2}{6\times 2}=\frac{10}{12}

\frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}

\implies \frac{5}{6}+\frac{1}{4}=\frac{10}{12}+\frac{3}{12}=\frac{13}{12}=1\frac{1}{12}

3) ∵ Pizza ate by Manny = \frac{1}{4}

Pizza ate by Frank = \farc{5}{8}

∴ Total pizza eaten = \frac{1}{4}+\frac{5}{8}

LCM(4, 8) = 8,

\frac{1}{4}=\frac{2}{8}

Thus, total pizza eaten =  \frac{2}{8}+\frac{5}{8}=\frac{7}{8}

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melisa1 [442]

The value of x that makes ABC~ RTS. is 12

<h3>How to determine the value of x?</h3>

From the figure, we have the following equivalent ratio:

2x + 6 : 40 = 36 : 48

Express as fraction

(2x + 6)/40 = 36/48

Multiply both sides by 40

2x + 6 = 30

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Divide both sides by 2

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Hence, the value of x is 12

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2 years ago
If the weight of the package is multiplied by 3/7 the result is 24. find the weight of the package
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W × ( 3 / 7 ) = 24 ;
w = ( 24 × 7 ) / 3 ;
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4 years ago
Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chord
emmainna [20.7K]

The arcs intersected by these chords are not congruent.

Given that two circles with different radii have chords AB and CD, such that AB is congruent to CD.

Let r₁ and r₂ be the radii of two different circles with centers O and O' respectively.

Assuming that the each of the ∠АОВ  and ∠CO'D is less than or equal to π.

Then, we have isosceles triangle AOB and CO'D such that,

AO = OB = r₁,

CO' = O'D = r₂,

Let us assume that r₁< r₂;

We can see that arc(AB) intersected by AB is greater than arc(CD), intersected by the chord CD;

arc(AB) > arc(CD)      .......(1)

Indeed,

arc(AB) = r₁ angle (AOB)

arc(CD) = r₂ angle (CO'D)

So, we have to prove that ;

∠AOB >∠CO'D       ......(2)

Since each angle is less than or equal to π, and so

∠AOB/2  and ∠CO'D/2 is less than or equal to π

it suffices to show that :

tan(AOB/2) >tan(CO'D/2) ......(3)

From triangle AOB :

tan(AOB/2) = AB/(2*r₁)

tan(CO'D/2) = CD/(2*r₂)

Since AB = CD and r₁ < r₂ (As obtained from the result of (3) ), therefore, arc(AB) > arc(CD).

Hence, for two circles with different radii have chords AB and CD, such that AB is congruent to CD but the arcs intersected by these chords are not congruent.

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scZoUnD [109]
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Answer:

1/4

Step-by-step explanation:

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