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Vilka [71]
3 years ago
13

What is the absolute value of negative 6

Mathematics
2 answers:
dimaraw [331]3 years ago
8 0
The absolute value of negative 6 is 2
antiseptic1488 [7]3 years ago
7 0
6 of course. (6 away from zero)
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10 and 40 I think maybe?

Step-by-step explanation:

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If f(x)=3x^2+1 and g(x) =1-x, what is the truly value of (f-g)(2)
Lilit [14]

Answer:

14

Step-by-step explanation:

f(x)=3x²+1

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Decompose 49 x 5 to<br> make it easier below<br> Explain
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Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()
lesantik [10]

Answer:

r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)

Step-by-step explanation:

A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then

e^0+c = 2

1+c = 2

c = 1

Thus, the first coordinate of r(t) is e^t + 1.

A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then

-cos(0)+c = 2

-1+c = 2

c = 3

Therefore, in the second coordinate r(t) is equal to -cos(t)+3.

Now, lets see the last coordinate.

A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that

2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0

Thus, c = 2

As a consecuence, the third coordinate of r(t) is tan(t) + 2.

As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).

6 0
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