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Dvinal [7]
3 years ago
5

If g(x) = 3x - 4, what is the value of g-(g(-2))? 100 X + 4 3 10

Mathematics
1 answer:
kozerog [31]3 years ago
3 0

-g(-2) is 10    

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

Step 1:

Given g(x) = 3x - 4. Find g(-2).

⇒ g(-2) = 3 × -2 - 4 = -6 - 4 = -10

Step 2:

Find -g(-2)

⇒ -g(-2) = -(-10) = 10

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Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, s=120\ ft/s

The height 'h' of the debris above the ground is given as:

h(t)=-16t^2+120t

As per question, h(t)=56\ ft. Therefore,

56=-16t^2+120t

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0

Using quadratic formula to solve for 't', we get:

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s

Therefore, the debris will reach a height of 56 ft twice.

When time t=0.5\ s during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at t=7\ s, the height is 56 ft.

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