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3 years ago

Solve for x (1/x) - (2x/(1+x^2)) = 0

2 answers:

7 0

(1/x) - (2x/(1+x^2)) = 0

Multiply by x(x^2+1)

1(x^2+1)+(−2x)(x)=0x(x2+1)

−x^2+1=0(Simplify both sides of the equation)

−x^2+1−1=0−1(Subtract 1 from both sides)

−x^2=−1

−x^2−1=−1−1(Divide both sides by -1)

x^2=1

x=√±1(Take square root)

x=1 or x=−1

padilas [110]3 years ago

6 0

$D:x\not=0 \wedge 1+x^2\not=0\\
D:x\not =0\\\\
\dfrac{1}{x}-\dfrac{2x}{1+x^2}=0|\cdot x(1+x^2)\\
1+x^2-2x^2=0\\
1-x^2=0\\
x^2=1\\
x=-1 \vee x=1


$

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Dafna1 [17]

The distance of the boat on the right to the finish line is 297.29 m and the distance of the boat on the left to the finish is 294.11 m, so the boat on the left will reach the finish line faster.

<h3>How do you know which boat will reach the finish line faster? </h3>

The image shows two triangles of which we know an angle and the value of the opposite leg. Therefore, we can perform a mathematical operation to know the value of the hypotenuse of these triangles, which would be the distance of the boats to the goal as shown below.

h = hypotenuse

Right Boat

Sen 22° = $\frac{110 m}{h}$

h = $\frac{110 m}{ Sin 22}$

h = $\frac{110 m}{0.37}$

h = 297,29 m

Left boat

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h = $\frac{100 m}{0.34}$

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According to the above, the distance of the boat on the left is less, so it can be inferred that this boat will reach the finish line faster.

Learn more about triangles in: brainly.com/question/2269348

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2 years ago

Compare the graphs of the function, ????(x) = −2(x + 3)^2 + 2 and ????(x) = 5(x + 3)^2 + 2. What do the graphs have

Luda [366]

Answer:

Similarity: Both functions has same vertex at point $(-3,2)$ .

Difference: $f(x)=-2(x+3)^2+2$ is downward opening parabola, while $g(x)=5(x+3)^2+2$ is an upward opening parabola.

Step-by-step explanation:

We have been given two functions as: $f(x)=-2(x+3)^2+2$ and $g(x)=5(x+3)^2+2$ . We are asked to compare the graphs of the given function.

We know that standard form of a parabola with vertex at point (h,k) is in format $y=a(x-h)^2+k$ .

Let us find the similarity in both graphs.

Upon comparing both functions with standard form of a parabola, we can see that vertex of the both parabolas is at point $(-3,2)$ .

Let us find the difference in both graphs.

We can see that leading coefficient of $g(x)=5(x+3)^2+2$ is positive, so it will be an upward opening parabola.

We can also see that leading coefficient of $f(x)=-2(x+3)^2+2$ is negative, so it will be a downward opening parabola.

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4 years ago

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algol [13]

You should make
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n200080 [17]

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slava [35]

Answer:

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