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3 years ago

What is the derivative of g(x)=e^(x^2+2x)+3x

Mathematics

2 answers:

igor_vitrenko [27]3 years ago

8 0

$g(x)=e^{x^2+2x}+3x\\
g'(x)=e^{x^2+2x}\cdot(2x+2)+3\\
g'(x)=2e^{x^2+2x}(x+1)+3
$

UNO [17]3 years ago

7 0

Ok first we can split it in two : $e^{x^2+2x}$ and $3x$ .

The derivative of $3x$ is 3.

For the first part, we use the chain rule : $[f(g(x))]'=g'(x)f'(g(x))$ hence $(e^{x^2+2x})'=(x^2+2x)'e^{x^2+2x}$ (since the derivative of the exponential is itself) hence $g'(x)=(2x+2)e^{x^2+2x}+3$

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Secant jkl and jmn are drawn to circle o from an external point ,j. if jk=8,lk=4 and jm=6 what is the length of jn answer

Archy [21]

See the picture attached to better understand the problem

we know that
If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
so
jl*jk=jn*jm------> jn=jl*jk/jm

we have
jk=8,lk=4 and jm=6
jl=8+4----> 12

jn=jl*jk/jm-----> jn=12*8/6----> jn=16

the answer is
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3 years ago

find the coordinates of the vertex of the following parabola algebraically. write your answer as an (x,y) point..y=x²+9

Alexandra [31]

Explanation:

using the parabola formula:

y = a(x-h)² + k²

vertex = (h, k)

We are given a parebola equation of: y = x²+9

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y = y

a = 1

(x-h)² = x²

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(x-h)² = (x + 0)²

h = 0

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1 year ago

Find the approximate circumference of a circle with a diameter of 100 yards.

elixir [45]

Answer:

D. 314 yds

Step-by-step explanation:

Given:

Diameter = 100 yds

Required;

Circumference of the circle

Solution:

Circumference of circle = πd

Plug in the value

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3 years ago

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Answer:

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nikdorinn [45]

Answer:

The equation does not have a real root in the interval $\rm [0,1]$

Step-by-step explanation:

We can make use of the intermediate value theorem.

The theorem states that if $f$ is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:

If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
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Of course, in our case, we will make use of the first one.

First, we need to proof that our function is continues in $\rm [0,1]$ , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval $\rm [0,1]$ , which means to evaluate the equation in 0 and 1:

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Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval $\rm [0,1]$ . I attached a plot of the equation in the interval $\rm [-2,2]$ where you can clearly observe how the graph does not cross the x-axis in the interval.

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