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3 years ago

Find f(g(4)). f(g(4)) =

Mathematics

2 answers:

RoseWind [281]3 years ago

8 0

Need more explanation

kondor19780726 [428]3 years ago

6 0

Answer:

Please attach the equations of f(x) and g(x) so we can solve.

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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))

sergejj [24]

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, f and g are inverses, then:

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

So, we can let j be the inverse function of h.

Function h is given by:

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

Find its inverse. Flip variables:

$x = 3y - 2$

Solve for y. Add:

$\displaystyle x + 2 = 3y$

Hence:

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

Therefore, a = 1/3 and b = 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:

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

3 0

2 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

4 0

2 years ago

Can please someone help me out and explain it step by step. Thank you

Anit [1.1K]

Answer: $x=7, \ y=-2$

Step-by-step explanation:

Given

Equations are $3x-y=23$ and $2x+3y=8$

take $3x-y=23$

$y=3x-23$

Put the value of y in another equation

$\Rightarrow 2x+3(3x-23)=8\\\Rightarrow 2x+9x-69=8\\\Rightarrow 11x=77\\\Rightarrow x=7$

Put the value of x in y

$y=3(7)-23=21-23=-2$

Hence, $x=7, \ y=-2$

7 0

2 years ago

If the area of a rectangle equals 28 square centimeters what is the height?

Slav-nsk [51]

So the length and width would be 7 and 4. hope that helped

3 0

3 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$

>> Therefore, your answers are x = -4 or x = 2.

Learn more about quadratic equations:

brainly.com/question/27638369

6 0

2 years ago

Read 2 more answers