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gtnhenbr [62]
2 years ago
10

(15-10)-(14+12+ (8+5)​

Mathematics
2 answers:
Travka [436]2 years ago
6 0

Answer:

-34

Step-by-step explanation:

Remove all brackets by multiplying the existing math signs then take it from there

nata0808 [166]2 years ago
5 0

(15-10)-(14+12+ (8+5)​

5 - (14 + 12 + (8 + 5)

5 - (14 + 12 + 13)

5 - 39

-34

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Can you help me find t value in this problem?
Eduardwww [97]

we have a maximum at t = 0, where the maximum is y = 30.

We have a minimum at t = -1 and t = 1, where the minimum is y = 20.

<h3>How to find the maximums and minimums?</h3>

These are given by the zeros of the first derivation.

In this case, the function is:

w(t) = 10t^4 - 20t^2 + 30.

The first derivation is:

w'(t) = 4*10t^3 - 2*20t

w'(t) = 40t^3 - 40t

The zeros are:

0 = 40t^3 - 40t

We can rewrite this as:

0 = t*(40t^2 - 40)

So one zero is at t = 0, the other two are given by:

0 = 40t^2 - 40

40/40 = t^2

±√1 = ±1 = t

So we have 3 roots:

t = -1, 0, 1

We can just evaluate the function in these 3 values to see which ones are maximums and minimums.

w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20

w(0) = 10*0^4 - 20*0^2 + 30    = 30

w(1) =  10*(1)^4 - 20*(1)^2 + 30 =  20

So we have a maximum at x = 0, where the maximum is y = 30.

We have a minimum at x = -1 and x = 1, where the minimum is y = 20.

If you want to learn more about maximization, you can read:

brainly.com/question/19819849

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