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3 years ago

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Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete

rs,19 imches and 50 inches,33 centimeters and 80 centimeters

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

$\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}$

b) 19 inches and 50 inches

$\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}$

c) 33 centimeters and 80 centimeters

$\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}$

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

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3 years ago

(-6x3-4x2-8)+(x3-x2-9x+4) add together and simply

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Step-by-step explanation:

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Add the like terms.

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3 years ago

Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

<em>So, the sides are: 38.5, 39.5, 40.5 and 41.5</em>

<em></em>

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

<em>So presently, Inzo is 15 years old and Miz is 10 years old</em>

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

5 0

2 years ago