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2 years ago

If f(1) = 1 and f(n + 1) = 5f(n) + 3 then find the value of f(4).

Mathematics

1 answer:

FrozenT [24]2 years ago

4 0

Answer:

218

Step-by-step explanation:

Because we need f(n) to calculate f(n+1), we have to calculate our values one step at a time. Starting with f(1) = 1,

f(2) = 5(f(1)) + 3 = 5(1) + 3 = 5+3 = 8

f(3) = 5(8) + 3 = 40 + 3 = 43

f(4) = 5(43) + 3 = 215 + 3 = 218

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aleksandrvk [35]

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2 years ago

What is the ratio of drivers to cars in a stock car race?

adoni [48]

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3 years ago

(-5, 3), (-1, 0), (3, -4), (-1, 2) pls find domain range and function

Nonamiya [84]

Answer:

Domain: {-5,-1,3} and Range: {3,0,-4,2}

Step-by-step explanation:

Just take my word for it please.

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2 years ago

Find f(g(x)). State the domain of the composite function.<br> f(x) = 3x-2/x+1<br> g(x) = x+5/2x-3

AnnyKZ [126]

Answer:

ℝ - {(-2/3),(3/2)}

Step-by-step explanation:

We want the domain of f(g(x)). So, firstly, we have to find the domain for g(x) and, then, for f(g(x)).

- Domain of g(x): Since the expression is a fracion, we must exclude the values of x that make null the denominator. Hence,

$g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}$

- Domain of f(g(x)): We'll find its expression:

$f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}$

Now, once again, we have to exclude the values of x that make the denominator equals to zero. Thus,

$f(g(x)) =\dfrac{21-x}{3x+2}\Longrightarrow 3x+2\neq0\iff \boxed{x\neq-\dfrac{2}{3}}$

Lastly, we may write the domanin of f(g(x)):

$D(f(g(x)) = \left]-\infty,-\dfrac{2}{3}\right[\cup\left]-\dfrac{2}{3},\dfrac{3}{2}\right[\cup\left]\dfrac{3}{2},\infty\right[$

or, just writing in a shorter way:

$\boxed{D(f(g(x)) = \mathbb{R}-\left\{-\dfrac{2}{3},\dfrac{3}{2}\right\}}$

7 0

2 years ago

Fish story: According to a report by the U.S. Fish and Wildlife Service, the mean length of six-year-old rainbow trout in the Ar

Yanka [14]

Answer:

11.29.

Step-by-step explanation:

We are given that The mean length of six-year-old rainbow trout in the Arolik River in Alaska is 478 millimeters with a standard deviation of 38 millimeters.

We are also given that these lengths are normally distributed.

So, The highest value is at mean .(Refer the attached figure).

Formula : $Percentile = \frac{Length}{Maximum} \times 100$

$54 = \frac{Length}{478} \times 100$

$\frac{5400}{478} =Length$

$11.297 =Length$

The 54th percentile of the lengths is 11.29.

4 0

3 years ago