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igomit [66]
3 years ago
8

A 208-long road is divided into 16 parts of equal length.Mr.Ward paints a 4-meter-long strip in each part.How many total meters

are not painted in the 208 meters of road?
Mathematics
2 answers:
Readme [11.4K]3 years ago
6 0
208 % 16 = 13
13 - 4 = meters not painted in a part = 9
9 × 16 (parts) = 144 meters of road not painted
Yuki888 [10]3 years ago
6 0

Answer

Find out the how many total meters are not painted in the 208 meters of road .

To prove

Let us assume that the total meters are not painted in the 208 meters of road be x.

As given

A 208-long road is divided into 16 parts of equal length.

Each\ road\ length\ when\ divided\ into\ 16\ equal\ parts = \frac{208}{16}

Each road length when divided into 16 equal parts = 13 meters

As given

Mr.Ward paints a 4-meter-long strip in each part.

Mr.Ward not paints  in each part =  13 - 4

                                                      = 9 meters

Thus

Total meters are not painted in the 208 meters of road = 9 × 16

                                                                                            = 144 meters

Therefore 144 meters are not painted in the 208 meters of road .




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If angle EFH(5x+1)’ angle HFG(62 degrees) and angle EFG (18x+11) find each measurement
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Answer:

∠EFH = 21°

∠HFG = 62°

∠EFG = 83°

Step-by-step explanation:

The diagram showing the angles has been attached to this response.

From the diagram, it can be deduced that;

Angle EFG = angle EFH + angle HFG

=> ∠EFG = ∠EFH + ∠HFG                -------------------(i)

From the question:

∠EFH = (5x + 1)°                    -------------(ii)

∠HFG = 62°                          -------------(iii)

∠EFG = (18x + 11)°                 -------------(iv)

<em>Substitute these values into equation (i) as follows;</em>

(18x + 11) = (5x + 1) + 62

=> 18x + 11 = 5x + 1 + 62

<em>Collect like terms and solve for x</em>

18x - 5x = 1 + 62 - 11

13x = 52

x = 4

Now, to get each measurement, substitute x = 4 into each of equations (ii) - (iv)

∠EFH = (5x + 1)°

∠EFH = (5(4) + 1)°

∠EFH = (20 + 1)°

∠EFH = 21°

∠HFG = 62°              [<em>Does not depend on x</em>]

∠EFG = (18x + 11)°

∠EFG = (18(4) + 11)°

∠EFG = (72 + 11)°

∠EFG = 83°

<u>Conclusion:</u>

∠EFH = 21°

∠HFG = 62°

∠EFG = 83°

8 0
3 years ago
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