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Westkost [7]
3 years ago
6

How does f(x) = 10x change over the interval from x = 5 to x = 6?

Mathematics
2 answers:
WINSTONCH [101]3 years ago
8 0

Answer:

f(x) increases by 900%

Step-by-step explanation:

f(x) increases by 900%

f(5) = 105

f(6) = 106

f(6)

f(5)

=  

106

105

= 106−5 = 10

Therefore, f(x) increases by a factor of 10 over the interval from x = 5 to x = 6.

Then,

A value increases by p% if it changes by a factor of 1 +  

p

100

.

f(6) = 10f(5)

f(6) = (1 + 9)f(5)

f(6) = (1 +  

900

100

)f(5)

f(6) = f(5) +  

900

100

f(5)

f(6) = f(5) + 900% · f(5)

Thus, f(6) is 900% larger than f(5). So, f(x) increases by 900% over the interval from x = 5 to x = 6.

kaheart [24]3 years ago
5 0

Answer:

C. 900 \%

Step-by-step explanation:

The given function is

f(x)=10^x

Substitute x=5 to get;

f(5)=10^5=100000

Substitute x=6 to get;

f(5)=10^6=1000000

The percentage increment of f(x) over the interval from x=5 to x=6 is

=\frac{100000-100000}{10000}\times100 \%

=\frac{900000}{10000}\times100 \%

=9\times100 \%

=900 \%

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6 0
3 years ago
For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, ????(????),
Effectus [21]

Answer:

a. Period, T = 12.57 hours. b. Latest time = 294.16 hours, 6 a.m 12 days later

Step-by-step explanation:

For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6sin(0.5t). (a) What is the period of the tides in hours? (b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?

a. The period, T of the tides is gotten from ω = 2π/T, where ω is the angular frequency of the wave and T its period. Comparing the sine part of equation of the tides with the general equqtion of a sine wave, Asinωt, thus Asinωt = 4.6sin(0.5t),

so ω = 0.5 rad/s.

Equating this to 2π/T implies

0.5 = 2π/T

therefore, T = 2π/0.5 = 12.566 hours ≈ 12.57 hours

b. The tide becomes too low if it is less than 2.5 m i.e d(t) < 2.5 m. So, equating d(t) as 2.5 in the tidal equation, we have d(t) = 5 + 4.6sin(0.5t).

2.5< 5+ 4.6sin(0.5t)

2.5-5 < 4.6sin(0.5t)

-2.5 < 4.6sin(0.5t)

-2.5/4.6 < sin(0.5t)

-0.543 < sin(0.5t)

sin⁻¹(-0.543)<0.5t

-32.92<0.5t. Since we cannot have negative time, we add 180 to -32.92 to give 147.08

147.08<0.5t

t>147.08/0.5=294.16 hours

So, if t is greater than 294.16 hours, the water will be shallow. That is, below 2.5 m. which is 12 days 6hours 8 min 38 s which is 6 a.m 12 days later

3 0
3 years ago
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