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2 years ago

Which function results after applying the sequence of transformations to f(x) = x^5

Mathematics

1 answer:

MaRussiya [10]2 years ago

4 0

Answer:

See explanation

Step-by-step explanation:

We want to find the function that results after $f(x)=x^5$ has gone through certain transformations.

If f(x) is shifted $a$ units to the right and b units up, the transformed function becomes;

$g(x)=(x-a)^5+b$

If f(x) is shifted $a$ units to the left and b units down, the transformed function becomes;

$h(x)=(x+a)^5-b$

If a reflection in the x-axis is part of the transformation then.

$g(x)=-(x-a)^5+b$

$h(x)=-(x+a)^5-b$

I hope this explanation will help you choose the correct answer from the original or complete question since you didn't specify the sequence of transformations that have been applied to f(x).

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

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Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

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$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

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2 years ago

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Answer:

Step-by-step explanation:

Units = 30

Thousands = 2,000

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2. Let's estimate the standard error, using the given formula, this way:

S.e = √ (0.46 * 0.54)/50 = √0.049 = 0.07

The correct answer is .07

3. Let's calculate the confidence limits of the 90% confidence interval, this way:

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