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3 years ago

A circle had a circumference of 138.16 meters. What is the diameter of the circle?

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

Divide 138.16 by 3.14. This gives a diameter of 44 meters.

Hope this helped :)

mixer [17]3 years ago

5 0

It's 44

C = 2 r pi

138.16 = 2 r pi

138.16 / 2 pi = r

r = 22 rounded to a whole number

r multiplied by 2 equals,44, which is the diameter

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Help mee A scout troop are hiking in a forest. Starting from their base, they walk 4.2km south followed by 7.1km west. They want

slega [8]

Answer:

They should walk on a bearing of 59.4 degrees

Step-by-step explanation:

Given

$South = 4.2km$

$West = 7.1km$

Required

The bearing back to the base

The given question is illustrated with the attached image.

To do this, we simply calculate the measure of angle a using:

$\tan(a) = \frac{Opposite}{Adjacent}$

$\tan(a) = \frac{7.1}{4,2}$

$\tan(a) = 1.6905$

Take arctan of both sides

$a = \tan^{-1}(1.6905)$

$a = 59.4^o$

4 0

3 years ago

The Social Justice Club is going to rent a mammoth barbeque to cook the hot dogs for the hot dog sale After contacting three ren

zimovet [89]

Answer:

(A)Cost of Rental A, C= 15x

Cost of Rental B, C=5x+50

Cost of Rental C, C=9x+20

(B)

i. Rental C ii. Rental A iii. Rental B

Step-by-step explanation:

Part 1

Let x be the number of hours of the barbeque use by the club.

Rental A: $15/h

Cost of Rental A, C= 15hx

Rental B: $5/h + 50

Cost of Rental B, C=5x+50

Rental C: $9/h + 20

Cost of Rental C, C=9x+20

Part 2

The graph of the three models is attached below

Part 3(11.05-4.30)

If the barbecue's usage hour is from 11.05 to 4.30 when the football match ends.

Number of Hours between 11.05am and 4.30pm=4 hours 25 Minutes = 4.42 Hours

Cost of Rental A, C= 15x=15(4.42)=$66.30

Cost of Rental B, C=5x+50 =5(4.42)+50=$72.10

Cost of Rental C, C=9x+20=9(4.42)+20=$59.78

Rental C should be chosen as it offers the lowest cost.

Part 4 (11.05-12.30)

Number of Hours = 12.30 -11.05 =1 hour 25 Minutes = 1.42 Hours

Cost of Rental A, C= 15x=15(1.42)=$21.30 Cost of Rental B, C=5x+50 =5(4.42)+50=$57.10 Cost of Rental C, C=9x+20=9(4.42)+20=$32.78

Rental A should be chosen since it offers the lowest cost.

Part 5

If the barbecue is returned the next day, say after 24 hours

Cost of Rental A, C= 15x=15(24)=$360 Cost of Rental B, C=5x+50 =5(24)+50=$170 Cost of Rental C, C=9x+20=9(24)+20=$236

Rental B should be chosen as it offers the lowest cost.

6 0

3 years ago

Explain how to use the break apart strategy to find 247+358

KIM [24]

1. Set up the long addition.
2 4 7
+3 5 8
_______

2. Calculate 7+8, which is 15.
since 15 is two-digit, we carry the first digit 1 to the next column.
  1
2 4 7
+ 3 5 8
________
   5

3. Calculate 4+5, which is 9. Now add the carry digit of 1, which is 10. Since 10 is two-digit, we carry the first digit 1 to the next column.
1 1
2 4 7
+ 3 5 8
________
0 5

4. Calculate 2+3, which is 5. Now add the carry digit of 1, which is 6.
1 1
2 4 7
+3 5 8
______
6 0 5

5. Therefore, 247 + 358 = 605.

605

8 0

4 years ago

Read 2 more answers

I need help with #18

Alexxx [7]

The ratio between the bigger triangle and the smaller triangle is 1:2, which means the smaller triangle is half the size of the bigger triangle.

You simply need to half all the values of the segments of the bigger triangle to find the values of the segments of the smaller triangle.

8/2 = x; 4 = x

6/2 = y; 3 = y

5 0

4 years ago

The graph below shows the average Valentine’s Day spending between 2003 and 2012

Lapatulllka [165]

The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

Part A.

From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

$\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year$

Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

$\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year$

Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

$\frac{103-102}{2010-2009} = \$1/year$

7 0

4 years ago