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Margarita [4]
3 years ago
10

Maurice and three other volunteers work at the hospital every week. if the number of hours each volunteer works is always the sa

me, how many total volunteer hours are worked per week at the hospital in order for each person to volunteer 12 hours a week?in the problem above, how many volunteers are working at this hospital?
Mathematics
2 answers:
agasfer [191]3 years ago
5 0
12 hours & 4 volunteers
matrenka [14]3 years ago
3 0

Answer:

48 hours ; 4 volunteers

Step-by-step explanation:

Maurice ( 1 volunteer) and three other volunteers = 4 total of volunteers working at the hospital. Each volunteer works always the same number of hours, and each volunteer works 12 hours per week.

4 volunteers X 12 hours a week = 48 volunteer hours a week.

48 is the total of volunteer hours the 4 volunteers work per week.

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Since 3 squared is 9 and 4 squared is 16, the square root of 13 will fall somewhere between 3 and 4
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3 years ago
Use the Angle Addition Postulate to find the measure of each angle.
Simora [160]
For number 15 and 16, you just have to find the absolute difference between the two points along the calibration of the protractor.

15. ∠BXC = |B - C| = |140° - 110°| = 30°
16. ∠BXE = |B - E| = |140° - 30°| = 110°

For numbers 20 and 21, apply the Angle Addition Postulate. This is when you add the individual interior angles to equate to the total angle.

20. ∠PQS = ∠PQR + ∠RQS
      112° = 72°+ 10x°
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21. ∠KLM = ∠KLN + ∠NLM
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5 0
3 years ago
Which of these four sets of side lengths will form a right triangle? Set 1 4 cm, 5 cm, 6 cm Set 2 8 in., 12 in., 20 in. Set 3 10
Alenkasestr [34]

Answer:  The correct option is (D) SET 4.

Step-by-step explanation:  We are to select the correct set of side lengths that will form a right-angled triangle.

To form a right-angled triangle, we must have the following relation:

<em>Perpendicular² + Base² = Hypotenuse².</em>

<em>Hypotenuse is the length of the largest side; perpendicular and base are the two legs of the triangle.</em>

SET 1 :  14 cm, 5 cm, 6 cm.  

We have

5^2+6^2=25+36=61,\\\\14^2=196.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 2 :  8 in., 12 in., 20 in.  

We have

8^2+12^2=64+144=208,\\\\20^2=400.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 3 :  10 mm, 20 mm, 30 mm.  

We have

10^2+20^2=100+400=500,\\\\30^2=900.

Therefore,

<em>Perpendicular² + Base² ≠ Hypotenuse².</em>

So, this set will not form a right-angled triangle.

SET 4 :  12 ft, 16 ft, 20 ft.  

We have

12^2+16^2=144+256=400,\\\\20^2=400.

Therefore,

<em>Perpendicular² + Base² = Hypotenuse².</em>

So, this set will form a right-angled triangle.

Thus, the SET 4 will form a right-angles triangle.

Option (D) is correct.

5 0
3 years ago
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B. Equation (hope this helps!)
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3 years ago
Pls help me asap. will give brainlist
Likurg_2 [28]

By pythagorus theorum In triangle ADC

AD²=AB²-BD²

AD²=(6.2)²-(2.7)²

AD=5.5

Again in triangle ADC

AC²=AD²+BC²

AC²=(5.5)²-(2.5)²

AC²=30.25-6.25

AC=4.8 approx..

6 0
2 years ago
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