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notsponge [240]
4 years ago
14

Regular exercise can help a person lower their resting heart rate. Suppose an individual’s resting heart rate, in beats per minu

te, is modeled by the function f(x) = –0.1x 3 + 0.6x 2 + 85 after x weeks of regular exercise.
How many weeks of exercise are necessary to lower the individual’s resting heart rate below 70 beats per minute?

7 weeks
8 weeks
9 weeks
10 weeks
Mathematics
2 answers:
netineya [11]4 years ago
8 0

Answer: 9 weeks

Step-by-step explanation:

-0.1³ + 0.6x² + 85 = 70

x ≈ 8.21998224

The answer is 9 weeks because you must round 8.22 up to 9. The reason you round up to 9 is because must include the numbers after the decimal in a week, but you cannot have 0.22 weeks, so 0.22 rounds up to 1.

sveta [45]4 years ago
5 0

Answer:

9

Step-by-step explanation:

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3 years ago
Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the
ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let r be the radius, d the diameter, and V the volume of the spherical balloon.

We know \frac{dV}{dt}=20 \:{in^3/s} and we want to find \frac{dr}{dt}

The volume of a spherical balloon is given by

V=\frac{4}{3} \pi r^3

Taking the derivative with respect of time of both sides gives

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

We now substitute the values we know and we solve for \frac{dr}{dt}:

d=2r\\\\r=\frac{d}{2}

r=\frac{12}{2}=6

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and \frac{dr}{dt} is:

\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

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Answer:

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Step-by-step explanation:

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3 years ago
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raketka [301]

Answer:

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Step-by-step explanation:

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Hello please help i’ll give brainliest
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Answer:

3rd Option

Step-by-step explanation:

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