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Harlamova29_29 [7]
3 years ago
8

15 m/s in km/h plz me bro​

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0
15m/s multiplied by 3.6 equal to 54km/h
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Four students were scheduled to give book reports in 1 hour. After the first report, 2/3 hour remained. The next two reports too
Novosadov [1.4K]

Answer:1/4 hour remaining

Step-by-step explanation:

Presenter 1 - 1/3 hour

Presenter 2 - 1/6 hour

Presenter 3 - 1/4 hour

Convert to least common denominator of 12

4/12 + 2/12 + 3/12 = 9/12

12/12 - 9/12 = 3/12

Reduce 3/12 to get 1/4 hour remaining

3 0
3 years ago
Paul is spreading grass seed on an area of his lawn that is 10 feet wide and 12 feet long.
Alex17521 [72]

Answer:

8 bags, but will have some extra.

Step-by-step explanation:

The area of the lawn is 120 square feet, and 15 x 8 =124.

5 0
3 years ago
A rectangular yard has a length that is 5 feet more than the width. Around the outside of the yard is a path made of bricks that
Phantasy [73]

Answer:

A) A = X^2 + 11X + 24

B) X = 18 ft

C) 414 ft^2

D) 132 ft^2

Step-by-step explanation:

A) Width W of yard and path = X + 3

Lenght L of path = X + 5 + 3 = X + 8 ft, therefore,

Area of yard and path = L x W = (X +3) x (X +8)

A = X^2 + 11X + 24

B) if area of the yard and path is 546 ft^2,

546 = X^2 + 11X + 24

X^2 +115X - 522 = 0 (quadratic equation)

Solving the quadratic equation gives

X = 18 and X = -29

Our answer can only be positive so we choose X = 18 ft

C) lenght of yard = 5 + 18 = 23 ft

Width = 18 ft

Therefore area = L x W = 23 x 18 = 414 ft^2

D) area of path = area of path and yard minus area of yard

= 546 - 414 = 132 ft^2

7 0
3 years ago
Determine which statements about the relationship are true. Choose two options.Each Friday, the school prints 400 copies of the
aliya0001 [1]

Answer:

It is c

Step-by-step explanation:

7 0
3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the given ven
Scrat [10]

9514 1404 393

Answer:

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

  AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

  BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

  CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, ...

  ∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

  ∠A = arccos(3/(4√130)) ≈ 86°

  ∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

  ∠B = arccos(49/(2√1118)) ≈ 43°

  ∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

  ∠C = arccos(37/(4√215)) ≈ 51°

The three angles are ...

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

_____

<em>Additional comment</em>

This sort of repetitive arithmetic is nicely done by a spreadsheet.

6 0
3 years ago
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