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fiasKO [112]
3 years ago
11

If g(x)= square root of 3-5x, find the domain of g'(x).

Mathematics
2 answers:
lys-0071 [83]3 years ago
7 0

Answer: Check the pic

Step-by-step explanation:

BaLLatris [955]3 years ago
5 0

Answer:

domain: x>3/5

Step-by-step explanation:

First we need to derive our function g(x) to get a new function g'(x)

To do this we will have to apply chain rule because we have an inner and outer functions.

Our G(x) = square root(3-5x)

Chain rule formula states that: d/dx(g(f(x)) = g'(f(x))f'(x)

where d/dx(g(f(x)) = g'(x)

g(x) is the outer function which is x^1/2

f(x) is our inner function which is 3-5x

therefore f'(x)= 1/2x^(-1/2) and f'(x) = -5

g'(f(x)) = -1/2(3-5x)^(-1/2)

Applying chain rule then g'(x) = 1/2 (3-5x)^(-/1/2)*(-5)

But the domain is the values of x where the function g'(x) is not defined

In this case it will be 3-5x > 0, because 3-5x is a denominator and anything divide by zero is infinity/undefined

which gives us x >3/5

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A 2-gallon bottle of fabric softener costs $16.64. What is the price per cup?
Burka [1]

Answer:

each cup would cost $0.52

Step-by-step explanation:

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2 years ago
Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
zloy xaker [14]

Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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Solve. 4.75t + 5 = 13.5t + 12 The solution is t = .
Eddi Din [679]
Answer: t= -0.799
This can also be written as t= -0.8

How to do it :

4.74t+ 5 = 13.5t + 12

first, subtract 4.74t from both sides
5= 8.76t + 12

next, subtract 12 from each side
-7 = 8.76t

next, divide 8.76 from each sides

-0.799 = t
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