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balandron [24]
2 years ago
7

Given f(x)=3x-1 and g(x)=2x-3, for which value of x does g(x)=f(2)?

Mathematics
2 answers:
zavuch27 [327]2 years ago
8 0

Answer: x=4

Step-by-step explanation:

Black_prince [1.1K]2 years ago
5 0
F(x) = 3x - 1
g(x) = 2x - 3
f(2) = 3(2) - 1 = 6 - 1 = 5

g(x) = f(2) => 2x - 3 = 5
2x = 5 + 3 = 8
x = 8/2 = 4
x = 4
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Can someone please help me with this logarithm equation!!
enot [183]

Answer:

\frac{629}{2}

Step-by-step explanation:

So, our equation is:

log_5(\frac{2x-4}{5})  = 3

In exponetial form, this looks like:

5^3=\frac{2x-4}{5}

Now lets cube the 5:

125=\frac{2x-4}{5}

Next, we can multiply the denominator on the right side by 5:

625=2x-4

We need to now get x alone by adding 4 to both sides:

629=2x

Finally, we divide by x's coefficent, 2, to get:

x=\frac{629}{2}

Hope this helps! :3

6 0
2 years ago
Draw a square that has (-2,4) and (3,-1) as two of its vertices
mr Goodwill [35]

Answer:

square with (-2,4) and (3,-1) vertices

3 0
3 years ago
Suppose a population of honey bees in Ephraim, UT has an initial population of 2300 and 12 years later the population reaches 13
Rama09 [41]

Answer:

common ratio: 1.155

rate of growth: 15.5 %

Step-by-step explanation:

The model for exponential growth of population P looks like: P(t)=P_i(1+r)^t

where P(t) is the population at time "t",

P_i is the initial (starting) population

(1+r) is the common ratio,

and r is the rate of growth

Therefore, in our case we can replace specific values in this expression (including population after 12 years, and  initial population), and solve for the unknown common ratio and its related rate of growth:

P(t)=P_i(1+r)^t\\13000=2300*(1+r)^{12}\\\frac{13000}{2300} = (1+r)^12\\\frac{130}{23} = (1+r)^{12}\\1+r=\sqrt[12]{\frac{130}{23} } =1.155273\\

This (1+r) is the common ratio, that we are asked to round to the nearest thousandth, so we use: 1.155

We are also asked to find the rate of increase (r), and to express it in percent form. Therefore we use the last equation shown above to solve for "r" and express tin percent form:

1+r=1.155273\\r=1.155273-1=0.155273

So, this number in percent form (and rounded to the nearest tenth as requested) is: 15.5 %

6 0
2 years ago
What is the constant term in the expansion of the binomial (2x + 3)3?
guajiro [1.7K]
The answer is 27 but I am not sure I hope my calculations ain't wrong
4 0
3 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
2 years ago
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