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zubka84 [21]
3 years ago
15

A store manager timed Janette to see how long it would take her to fold and put away a sweater, a shirt, a pair of pants, and a

scarf. It took her 26.1 seconds for the shirt, 24.3 seconds for the sweater, 32.8 seconds for the pants, and 18.2 seconds for the scarf. What was the average time it took Janette to fold and put away all four items?
Mathematics
2 answers:
Tasya [4]3 years ago
7 0
Add them all up and divide by 4
101.4/4 = 25.35 or 25.4 seconds
ArbitrLikvidat [17]3 years ago
7 0

Answer:

25.4

Step-by-step explanation:

thats the answer... your welcome

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Use the given measures of the triangles to write the indicated trigonometric value. Then find the measure of the indicated angle
Alja [10]

Answer: read the comments for extra added more accurate information this was in the comments and by HOZI3RSMUS3 so it’s not mine


We're suppose to measure the trigonometric values of the triangles and the values of indicates angles .For that,We need to know that,Sin θ = side opposite to angle  θ / HypotenuseCos θ = side adjacent to angle θ / HypotenuseTan  θ = side opposite to angle θ/side adjacent to angle θNow,In the first triangle,We're provided by,side opposite to angle  θ =QR =24side adjacent to angle θ = PR = 10Hypotenuse = PQ = 26So,Sin P = 24/26Cos P = 10/26Tan P = 24/10m∠P  = sin p° = 24/26m∠P= sin^-1(24/26)m∠P= 67.38 °In the second triangle,We're provided by,side opposite to angle  θ =MO =9side adjacent to angle θ = NO = 40Hypotenuse = MN= 41So,Sin N= 9/41Cos N= 40/41Tan N= 9/40m∠N  = sin N° = 9/41m∠N= sin^-1(9/41)m∠N= 12.68°In the third triangle,We're provided by,side opposite to angle  θ =AC =15side adjacent to angle θ = AB = 8Hypotenuse = BC = 17So,Sin B = 15/17Cos B = 8/17Tan B = 15/8m∠B  = sin p° = 15/17m∠B= sin^-1(15/17)m∠B= 61.93°In the fourth triangle,We're provided by,side opposite to angle  θ =BC =8side adjacent to angle θ = AC = 6Hypotenuse = AB = 10So,Sin A = 8/10Cos A = 6/10Tan A = 8/6m∠A  = sin p° = 8/10m∠A= sin^-1(8/10)m∠P= 53. °*53.13°

Step-by-step explanation:

4 0
2 years ago
Mr. Clarke built a deck around the swimming pool and sandbox in his backyard. What is the area of the decking that surrounds the
ss7ja [257]

Answer:

1,174.8\ ft^{2}

Step-by-step explanation:

we know that

The area of the decking is equal to the area of the parallelogram minus the area of the pool minus the area of the sand box

Remember that

21\frac{1}{2}\ ft=21.5\ ft

so

A=(50)(35)-(20)(25)-(21.5)(7)/2\\A= 1,750-500-75.25\\A=1,174.8\ ft^{2}

4 0
3 years ago
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