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LiRa [457]
3 years ago
9

which statements must be true check all that apply consider the reflection of triangle ABC across the line of reflection line PT

Mathematics
1 answer:
melomori [17]3 years ago
3 0

2nd, 3rd, 5th, & 6th are the correct answers.

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Find f. (Use C for the constant of the first antiderivative and D for the constant of the second antiderivative.)
sashaice [31]

Answer:

f(x) = x^3  -sinx +Cx+D

Step-by-step explanation:

Given that:

f ''(x)= 6x +sinx

We are given the 2nd derivative of a function f(x) and we need to find f(x) from that.

We will have to integrate it twice to find the value of f(x).

Let us have a look at the basic formula of integration that we will use in the solution:

1.\ \int {(a\pm b)} \, dx =\int {a} \, dx + \int {b} \, dx \\2.\ \int {x^n} \, dx = \dfrac{x^{n+1}}{n+1}+C\\3.\ \int {sinx} \, dx = -cosx+C\\4.\ \int {cosx} \, dx = sinx+C

\int\ {f''(x)} \, dx =\int\ {(6x +sinx)} \, dx \\\Rightarrow \int\ {6x} \, dx  + \int\ {sinx} \, dx \\\\\Rightarrow 6\dfrac{x^2}{2} -cosx +C\\\Rightarrow 3{x^2} -cosx +C\\\Rightarrow f'(x)=3{x^2} -cosx +C\\

Now, integrating it again to find f(x):

f(x) =\int {f'(x)} \, dx =\int{(3{x^2} -cosx +C)} \, dx \\\Rightarrow \int{3{x^2}} \, dx  -\int{cosx} \, dx  +\int{C} \, dx\\\Rightarrow 3\times \dfrac{x^3}{3}  -sinx +Cx+D\\\Rightarrow x^3  -sinx +Cx+D\\\\\therefore f(x) = x^3  -sinx +Cx+D

5 0
4 years ago
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