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neonofarm [45]
3 years ago
13

Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))

Mathematics
1 answer:
sergejj [24]3 years ago
3 0

Answer:

\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, <em>f</em> and <em>g</em> are inverses, then:

\displaystyle f(g(x)) = g(f(x)) = x

So, we can let <em>j</em> be the inverse function of <em>h</em>.

Function <em>h</em> is given by:

\displaystyle h(x) = y = 3x-2

Find its inverse. Flip variables:

x = 3y - 2

Solve for <em>y. </em>Add:

\displaystyle x + 2 = 3y

Hence:

\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}

Therefore, <em>a</em> = 1/3 and <em>b</em> = 2/3.

We can verify our solution:

\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}

And:

\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x  \stackrel{\checkmark}{=} x\end{aligned}

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<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀          where

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Answer:

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Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shaded area is equal to the area of the square less the area not shaded.

There are 4 "not shaded" regions.

step 1

Find the area of square ABCD

The area of square is equal to

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we have

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and the area of 4 "not-shaded" regions is:

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step 3

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Remember that the area of the shaded region is the area of the square less 4 "not shaded" regions:

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