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sdas [7]
2 years ago
9

Emma, Brandy, and Damian will cut a rope that is 29.8 feet long into 3 jump ropes. Each of the three jump ropes will be the same

length. Write a division sentence using compatible numbers to estimate the length of each rope
Mathematics
1 answer:
Vesnalui [34]2 years ago
3 0

Answer:

The length of each rope is 9.93 feet (Approx).

Step-by-step explanation:

As given

Emma, Brandy, and Damian will cut a rope that is 29.8 feet long into 3 jump ropes.

Each of the three jump ropes will be the same length.

Thus

Length\ of\ each\ rope = \frac{29.8}{3}

Length\ of\ each\ rope = 9.93\ feet(Approx)

Therefore the length of the each rope is 9.93 feet (Approx).



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Answer:

It will take 12 weeks to pay for the DVD players.

Step-by-step explanation:

Given in the question that,

total cost of DVD bought by Adela = $210

Step 1

Adela makes a down payment of $30

$210 - $30 = $180

Remaining cost = $180

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Adela pays $15 each week

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7 0
2 years ago
Una fabrica produce 132 litros de yogurt diarios .Con 49 litros se llenan botellas de 0,25litros xada una y con el resto que que
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Answer:

z = 362\,botellas

Step-by-step explanation:

(This exercise has been presented in Spanish and for that reason explanation will be held in Spanish)

La cantidad de botellas de 0,25 litros que se llenan con 49 litros de yogurt es:

x = \frac{49\,L}{0.25\,\frac{L}{botella} }

x = 196\,botellas

La cantidad remanente de yogurt es:

V_{r} = 132\,litros - 49\,litros

V_{r} = 83\,litros

Ahora, la cantidad de botellas de 0,5 litros que se llenan con el volumen remanente es:

y = \frac{83\,litros}{0.5\,\frac{litros}{botella} }

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Finalmente, el total de botellas a llenar es:

z = 196\,botellas + 166\,botellas

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2 years ago
Central City High School's robotics team is at a competition. In the last round, the team earned 80 points for their robot climb
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344=12x+80
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Lori creates the following design for a T-shirt.
kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7


Area of rectangle 1 = 8.68 sq.in

\Rightarrow FH \times HI=8.68

\Rightarrow EB \times HI=8.68  (∵ sides of the rectangle 2)

\Rightarrow 2.8 \times HI=8.68

\Rightarrow HI=3.1


Area of Triangle 1 = 6.48 sq.in

\Rightarrow \frac{1}{2}\times AE \times EG= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48  (∵ sides of the rectangle 1)

\Rightarrow EF+3.1= 5.4

\Rightarrow EF=2.3

\Rightarrow BH=2.3  (∵ sides of the rectangle 2)


BC = BH+HI+IC

\Rightarrow 11.7= 2.3+3.1+IC

\Rightarrow IC=6.3


The area of Rectangle 2,

=EB\times BH =2.8\times 2.3=6.44\ sq.in


The area of Triangle 2,

\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in


The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in


6 0
2 years ago
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