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Greeley [361]
3 years ago
13

The increase in pressure (P) is a linear function of the depth (d) to which a scuba diver descends. This is the function: P(d)=0

.445d + 14.7.
1.) What does the initial value represent?
2.) What does the rate of change represent?​
Mathematics
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

Initial value = 14.7

Rate of change = 0.445

Step-by-step explanation:

Given the linear function :

P(d) = 0.445d + 14.7

The linear equation above takes the general. Form of a slope - intercept equation :

y = mx + c

Where ; m = slope or gradient = rate of change in y per unit change in x

c = intercept , which is the intial value of y when x = 0

Thus using the above information ;

Initial value = intercept value = 14.7

The rate of change is the slope value = 0.445

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Maurinko [17]
The two numbers are 2 and 15
7 0
2 years ago
Read 2 more answers
Pls pls help it is due todayyy
Pani-rosa [81]

Answer:

I did this today and its H

Step-by-step explanation:

Please lmk if im wrong, but im sure its right but js ;)

6 0
3 years ago
LCM and the GCF of 12 and 32?
OverLord2011 [107]
The greatest common factor of 12 and 32 is 4.

The least common multiple of 12 and 32 is 96.

Hope I helped☺☺
7 0
4 years ago
If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams
Juliette [100K]

Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

N=N_0e^{-\lambda t}

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34, t= 10

N=N_0e^{-\lambda t}

\Rightarrow \frac {N}{N_0}=e^{-\lambda t}

\Rightarrow \frac34 =e^{-\lambda .10}

Taking ln both sides

\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|

\Rightarrow ln|\frac34|=-10\lambda

\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}

Now , N₀= 512 grams, N= 121.5 grams, t=?

N=N_0e^{-\lambda t}

\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}

\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}

\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}

Taking ln both sides

\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|

\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}

\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}

\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

8 0
4 years ago
Find the inverse of P(x)=-15x^2+350x-2000 and state it’s domain.
natima [27]

Answer:

The function given is:

P(x) = 15x² + 350x - 2000

or

y = 15x² + 350x - 2000

Interchange x and y:

x = 15y² + 350y - 2000

x = (y√15)² + 2(y√15)(45.19) + (45.19)² - (45.19)² - 2000

x = ( y√15 + 45.19)² - 4042.14

( y√15 + 45.19)² = x + 4042.14

Take square root:

y√15 + 45.19 = (√(x + 4042.14))

y√15 =  (√(x + 4042.14)) - 45.19

y = \sqrt{\frac{x+4042.14}{15}}-11.67

f^{-1}(x) = \sqrt{\frac{x+4042.14}{15}}-11.67

The minimum value of x, for which function is defined and real, is x= -4042.14

Hence, the range of inverse is x ≥ -4042.14

4 0
3 years ago
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