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2 years ago

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If f(x) = x2 – 1 and g(x) = 2x – 3, what is the domain of (fºg)(x)?

1 answer:

RUDIKE [14]2 years ago

5 0

Answer:

(-∞, ∞)

Step-by-step explanation:

Given,

$f(x)=x^2-1 \ and\ g(x)=2x-3$

Then (fºg)(x) would be ,

$(2x-3)^2-1\\4x^2-12x+9-1\\4x^2-12x+8$

Since this is a polynomial, we know domain of a polynomial is all real number.

Therefore the domain of this (fºg)(x) is (-∞, ∞).

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Someone solve this for me please: Find a vector v such that || v || = 10 and v is in same direction as 2i + 3j

drek231 [11]

Answer:

Step-by-step explanation:

$\overrightarrow{a}=2*\overrightarrow{i}+3*\overrightarrow{j}\\\\||\overrightarrow{a}||=\sqrt{2^2+3^2} =\sqrt{13} \\\\\boxed{\overrightarrow{v}=\dfrac{20}{\sqrt{13}}*\overrightarrow{i}+\dfrac{30}{\sqrt{13}}*\overrightarrow{j}} \\\\\\||\overrightarrow{a}||=\sqrt{(\dfrac{20}{\sqrt{13}})^2+(\dfrac{30}{\sqrt{13}})^2} =\sqrt{\dfrac{400+900}{13} }==\sqrt{\dfrac{1300}{13} }=\sqrt{100}=10 \\$

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2 years ago

The path of a skateboarder can be modeled by the equation (x – 3)2 = –8(y – 8), where distance is in feet and air resistance is

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The maximum height and horizontal distance is required.

Maximum height 8 feet; horizontal distance 16 feet

The given path is $(x-3)^2=-8(y-8)$

If the skateboarder jumps at x = 0 she will land at y = 0. However there will be at horizontal displacement.

$(x-3)^2=-8(0-8)\\\Rightarrow (x-3)^2=8^2\\\Rightarrow x-3=\pm 8\\\Rightarrow x-3=8\\\Rightarrow x=11$

or

$x-3=-8\\\Rightarrow x=-5$

The distance between -5 and 11 is $11-(-5)=16$

Middle point would be $\dfrac{16}{2}=8$ from $-5$ .

$-5+8=3$

So, $x=3$

The maximum height will be

$(3-3)^2=-8(y-8)\\\Rightarrow y-8=0\\\Rightarrow y=8$

The maximum height is 8 feet and the horizontal distance is 16 feet.

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Solve the system useing elimination 4x+8y=16 6x-8y=4

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Answer:

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Step-by-step explanation:

Add the two equations to remove "y" and solve for x:

4x+8y=16

6x-8y=4

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10x=20

x=2

Plug x=2 into one of the original equations to get the value of "y":

4x+8y=16

4(2)+8y=16

8+8y=16

8y=8

y=1

Therefore, x=2 and y=1

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2 years ago

A train whistle is 580 Hz when stationary. It is coming toward you at 18.8 m/s. What frequency do you hear?

Usimov [2.4K]

Answer:

612 Hz.

Step-by-step explanation:

The frequency of the sound will be proportionally higher when coming towards you. The increase is related to the speed of the train.

So the frequency will be:

580 + (580 / 343) * 18.8

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