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3 years ago

Which graph represents a reflection of f(x) = 1/10 (10)x across the y-axis?

Mathematics

2 answers:

nekit [7.7K]3 years ago

7 0

This is the concept of transformations, given the function given by f(x)=1/10(10)^x, when the function is reflected along the y-axis the new function will be given by f(x)=1/10(10)^(-x). This will mirror the original function.

Nesterboy [21]3 years ago

3 0

Answer:

We have to find a graph which represents a reflection of f(x) = 1/10 (10)x across the y-axis.

As we know that on reflecting the graph across the y-axis the x-coordinate of the point changes to opposite sign nut with same magnitude and the y-coordinate remains same.

Hence, the function f(x) is transformed to:

$f(x)=\dfrac{1}{10}\times 10^{-x}$

The graph of the function passes through the point (0,0.1) and is a strictly decreasing function.

and the end behavior of the graph is that:

when x→∞ f(x)→ 0

when x→ -∞ f(x) → ∞

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Can someone please help?

noname [10]

Answer:

Step-by-step explanation:

this one is correct

1/5+1/5=2/5

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3 years ago

Solve using Pythagoras Theorem (please)

aleksklad [387]

Answer:

HA = 16.2 m

DE = 17 m

Step-by-step explanation:

From the base of the cuboid, HDA will form a right angle triangle, where;

DA = 15 m

HA = 6 m

HA is the hypotenuse

Using pythagoras theorem;

HA = √(15² + 6²)

HA = √(225 + 36)

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HA = 16.155 m

Approximating to 1 decimal place gives;

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3 years ago

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

harina [27]

Answer:

The volume of the solid is 714.887 units³

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: V = $\int\limits^a_b {2\pi xf(x)} \, dx$ ,

where 2πx · f(x) is the surface area of the cylinder shell and

dx is its thickness

* Lets solve the problem

∵ y = $x^{\frac{5}{2}}$

∵ The plane region is revolving about the y-axis

∵ y = 32 and x = 0

- Lets find the volume by the shell method

- The definite integral are x = 0 and the value of x when y = 32

- Lets find the value of x when y = 0

∵ $y = x^{\frac{5}{2}}$

∵ y = 32

∴ $32=x^{\frac{5}{2}}$

- We will use this rule to find x, if $x^{\frac{a}{b}}=c, then=== x=c^{\frac{b}{a}}$ , where c

is a constant

∴ $x=(32)^{\frac{2}{5}}=4$

∴ The definite integral are x = 0 , x = 4

- Now we will use the rule

∵ $V = \int\limits^a_b {2\pi}xf(x) \, dx$

∵ y = f(x) = x^(5/2) , a = 4 , b = 0

∴ $V=2\pi \int\limits^4_0 {x}.x^{\frac{5}{2}}\, dx$

- simplify x(x^5/2) by adding their power

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{7}{2}}} \, dx$

- The rule of integration of $x^{n} is ==== \frac{x^{n+1}}{(n+1)}$

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{9}{2}}} \, dx=2\pi[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}]$ from x = 0 to x = 4

∴ $V=2\pi[\frac{2}{9}x^{\frac{9}{2}}]$ from x = 0 to x = 4

- Substitute x = 4 and x = 0

∴ $V=2\pi[\frac{2}{9}(4)^{\frac{9}{2}}-\frac{2}{9}(0)^{\frac{9}{2}}}]=2\pi[\frac{1024}{9}-0]$

∴ $V=\frac{2048}{9}\pi=714.887$

* The volume of the solid is 714.887 units³

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Answer:

Step-by-step explanation:

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