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aleksandr82 [10.1K]
4 years ago
10

The answer to the problem

Mathematics
1 answer:
andreev551 [17]4 years ago
8 0

Answer:100

Step-by-step explanation:

52+28=80

180-80=100


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3 years ago
If F(x)=integral of sqrt(t^3+1) dt from 0 to x, then F'(2)=?
Arlecino [84]
To answer this question, first
let ∫ √(t³+1) dt = g(t) + C 
<span>Then g'(t) = √(t³+1) </span>

<span>F(x) = ∫₀ˣ √(t³+1) dt = g(t) |₀ˣ = g(x) - g(0) </span>

<span>Now g(x) is some function of x, while g(0) is a constant </span>

<span>F(x) = g(x) - g(0) </span>

<span>Differentiate both sides: </span>
<span>F'(x) = g'(x) - 0 = √(x³+1) </span>

<span>So you are correct, in this case, we simply replace t with x (this is not always the case) </span>

<span>F'(2) = √(2³+1) = √9 = 3 </span>

<span>You MUST remember that when dealing with square roots, we have: </span>
<span>x² = 4 -----> x = -2 or 2 </span>
<span>x = √4 ----> x = 2 </span>

<span>That's why in the quadratic formula: x = (-b ± √(b²-4ac)) / (2a), we use a ± sign in front of square root, otherwise, if we could willy-nilly assign positive and negatives value to √(b²-4ac), then we would have no need for the ± sign. </span>

<span>Also, when solving x² = 4, we usually have intermediate step </span>
<span>x = ± √4, where +√4 (or simply √4) = 2, and -√4 = 2
</span>
7 0
3 years ago
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