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aleksandr82 [10.1K]
3 years ago
5

Ou decide to ride your bike for 30 minutes, raising your heart rate to the middle of your target heart rate zone. What type of a

ctivity is this an example of?
Aerobic, because it uses strong bursts of energy for a short duration
Aerobic, because it uses oxygen for a long duration
Anaerobic, because it uses oxygen for a long duration
Anaerobic, because it uses strong bursts of energy for a short duration
Mathematics
2 answers:
mina [271]3 years ago
6 0
Aerobic, it uses oxygen for a longer period of time.

CaHeK987 [17]3 years ago
3 0

This activity is an aerobic activity. This is known as aerobic because this consists of cardio exercises. During these activities, the heart rate and breathing rate will increase. These require free oxygen for a longer time.

Hence, Aerobic, because it uses oxygen for a long duration- is the correct answer.

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For number 2 I think it’s B but I am not sure

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3 years ago
In a ten pin bowling the highest possible score in a single game is 300. At one point in the season, John had an average score o
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Answer:

John need to get 293 in his next game to bring his average up to 183.

Step-by-step explanation:

John starts with N games played and averaging 177 points per game.

In game N+1, John scored 199 points.

\frac{177N + 199}{N+1}=178\\177N+199=178N+178\\N=21\\

So, John has now played (N+1) = 22 games.

We need to know what score John needs in game 23 to bring his average to 183.

\frac{177*21 + 199+Y}{23}=183\\3916+Y=4209\\Y=293\\

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7 0
4 years ago
The diagram shows the sector of a circle with the centre O and radius 6cm.
cricket20 [7]

Answer:

Area of the shaded region = 1.92 cm²

Step-by-step explanation:

From the picture attached,

Area of the shaded region = Area of the sector OMN - Area of the triangle OMN

Area of sector OMN = \frac{\theta}{360}(\pi r^{2})

Here, θ = Central angle of the sector

r = radius of the sector

Area of sector OMN = \frac{50}{360}(\pi )(6)^2

                                  = 15.708 square cm

Area of ΔOMN = 2(ΔOPN)

Area of ΔOPN = \frac{1}{2}(OP)(PN)

Area of ΔOMN = OP × PN

In ΔOPN,

sin(25°) = \frac{PN}{ON}

PN = ONsin(25°)

     = 6sin(25°)

     = 2.536 cm²

cos(25°) = \frac{OP}{ON}

OP = ONcos(25°)

OP = 6cos(25°)

OP = 5.438 cm

Area of ΔOMN = 2.536 × 5.438

                         = 13.791 cm²

Area of the shaded region = 15.708 - 13.791 = 1.917 cm²

                                             ≈ 1.92 cm²

3 0
3 years ago
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