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Alekssandra [29.7K]
3 years ago
14

For a craft activity at a day care, each child will need 1.75 yards of ribbon and 2.8 yards of fabric. There are 25 yards of rib

bon and 30 yards of fabric available. Estimate the number of children that can participate in the activity. Explain your reasoning.
Mathematics
1 answer:
Kamila [148]3 years ago
7 0

Total number of yards of ribbon available =  25 yards.

Number of yards of fabric available  = 30 yards.

Number of yards of ribbon a child need  = 1.75 yards

Number of yards of fabric a child need = 2.8 yards.

<em>Total number of children participate in the activity for 25 yards of ribbon = 25/1.75 = 14.28 approximately 15 children.</em>

<em>Total number of children participate in the activity for 30 yards of fabric = 30/2.8 = 10.7 approximately 11 children.</em>

<h3>Therefore, maximum 11 children can participate in the activity because each child required both 1.75 yards of ribbon and 2.8 yards of fabric .</h3>
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An object is traveling at a steady speed of 10 and one tenth km​/h. How long will it take the object to travel 4 and nine tenths
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Answer:

Estimated Answer=\frac{1}{2} hour

Exact Answer= \frac{49}{101} hr

Step-by-step explanation:

Steady speed of the object =10 and one tenth km​/h = 10\frac{1}{10} km/hr

Distance to be Covered=4 and nine tenths km =4\frac{9}{10} km

To the nearest integer,

Steady speed of the object = 10 km/hr

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Time Taken = \frac{Distance}{Average Speed}

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The estimated answer is 1/2 hr.

Next, we determine the exact answer.

Time Taken = \frac{Distance}{Average Speed}\\  = 4\frac{9}{10} \div 10\frac{1}{10}\\=\frac{49}{10} \div \frac{101}{10}\\=\frac{49}{10} X \frac{10}{101}\\=\frac{49}{101} hr

The exact time taken is \frac{49}{101} hr

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3 years ago
Evaluate the expression.
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Answer:

C

Step-by-step explanation:

The rest are correlation

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Answer:

Step-by-step explanation:

Triangle 1   given:  100° , 17 yds,  30 yds

Triangle 2 given: 30 ,17,18

law of cosines: c^2 = a^2 + b^2 − 2ab cos(C)

we'll also  need the formula for a SSS triangle

Cos(C) = a^2 + b^2 - c^2 / 2ab

keeping the above in mind we can now solve both triangles

c = AC for the first triangle

solve law of cosines for what we know

c^2= 17^2 + 30^2 - 2*17*30*Cos(100)

c = 36.96

now use SSS triangle  formula to find the angle C, but c = 17 now , not 36.96, that's b

C = arcCos(17^2+36.96^2-30^2/ 2*17*36.96)

C = 42.24°

now use 180 = 100 +42.24 +A   to solve the last angle

A= 37.75°

Triangle 1 solved

A= 37.75°

B= 100

C = 42.24°

a= 17

b=30

c=36.96

Triangle 2  

given a=17 , b= 18 , c =30

use SSS formula to find one angle, which ever one you want in this case

Cos(C) = a^2 + b^2 - c^2 / 2ab

C = arcCos(17^2+18^2-30^2 / 2*17*18

C= 117.966°

solve for A now

Cos(A) =b^2+c^2-a^2 / 2bc

A = arcCos(18^2+30^2-17^2/2*18*30)

A = 30.03

use 180 again to solve for last angle

180 = 117.966+30.03+B

32.001 = B

Triangle 2 solved

A = 30.03°

B= 32.001°

C= 117.966°

a=17

b=18

c=30

:P  whew  

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3 years ago
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