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kifflom [539]
2 years ago
14

For which pairs of function is (f x g ) (x) = 12x

Mathematics
2 answers:
Helga [31]2 years ago
6 0

Answer:

d f(x) = 4 and g(x) =12x

Step-by-step explanation:

just took the quiz

Contact [7]2 years ago
4 0
The answer is (f * g) (x) = 12x
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2 years ago
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What two numbers multiply to 11 and add to -16
VladimirAG [237]
They're irrational numbers, so they can't be exactly written down.
Rounded to the nearest thousandth, they are

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5 0
3 years ago
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Could you please look at this question?
MrMuchimi

Answer:

  a. see attached

  b. H(t) = 12 -10cos(πt/10)

  c. H(16) ≈ 8.91 m

Step-by-step explanation:

<h3>a.</h3>

The cosine function has its extreme (positive) value when its argument is 0, so we like to use that function for circular motion problems that have an extreme value at t=0. The midline of the function needs to be adjusted upward from 0 to a value that is 2 m more than the 10 m radius. The amplitude of the function will be the 10 m radius. The period of the function is 20 seconds, so the cosine function will be scaled so that one full period is completed at t=20. That is, the argument of the cosine will be 2π(t/20) = πt/10.

The function describing the height will be ...

  H(t) = 12 -10cos(πt/10)

The graph of it is attached.

__

<h3>b. </h3>

See part a.

__

<h3>c.</h3>

The wheel will reach the top of its travel after 1/2 of its period, or t=10. Then 6 seconds later is t=16.

  H(16) = 12 -10cos(π(16/10) = 12 -10cos(1.6π) ≈ 12 -10(0.309017) ≈ 8.90983

The height of the rider 6 seconds after passing the top will be about 8.91 m.

8 0
2 years ago
Which function has a minimum and is transformed to the right and down from the parent function, f(x) = x2? g(x) = –9(x2 + 2x + 1
nata0808 [166]
D. Last Answer is the correct answer to your question
8 0
3 years ago
What are the solutions of the inequality? -2(3x + 2) ≥ -6x - 4 Question 2 options: all real numbers x ≤ 6 no solution x ≥ 0
Tju [1.3M]

Solution, -2\left(3x+2\right)\ge \:-6x-4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:True\quad \forall \:x\in \mathbb{R}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}

Steps:

-2\left(3x+2\right)\ge \:-6x-4

\mathrm{Expand\:}-2\left(3x+2\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-2,\:b=3x,\:c=2,\\-2\cdot \:3x+\left(-2\right)\cdot \:2,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-2\cdot \:3x-2\cdot \:2,\\\mathrm{Simplify}\:-2\cdot \:3x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6,\\-6x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4,\\-6x-4,\\-6x-4\ge \:-6x-4

\mathrm{Add\:}4\mathrm{\:to\:both\:sides},\\-6x-4+4\ge \:-6x-4+4

\mathrm{Simplify},\\-6x\ge \:-6x

\mathrm{Add\:}6x\mathrm{\:to\:both\:sides},\\-6x+6x\ge \:-6x+6x

\mathrm{Simplify},\\0\ge \:0

\mathrm{Therefore,\:the\:final\:solution\:is},\\True\quad \forall \:x\in \mathbb{R}

Or\:No\:Solution\:x\geq 0


8 0
2 years ago
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