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docker41 [41]
3 years ago
7

Use the information below to complete the problem: p(x) = (1)/(x + 1)

Mathematics
1 answer:
bija089 [108]3 years ago
8 0

Given:

The functions are:

p(x)=\dfrac{1}{x+1}

q(x)=\dfrac{1}{x-1}

To find:

The rational expression for p(x)-q(x).

Solution:

We have,

p(x)=\dfrac{1}{x+1}

q(x)=\dfrac{1}{x-1}

Now,

p(x)-q(x)=\dfrac{1}{x+1}-\dfrac{1}{x-1}

p(x)-q(x)=\dfrac{(x-1)-(x+1)}{(x+1)(x-1)}

p(x)-q(x)=\dfrac{x-1-x-1}{x^2-1^2}             [\because a^2-b^2=(a-b)(a+b)]

p(x)-q(x)=\dfrac{-2}{x^2-1}

Therefore, the required rational expression for p(x)-q(x) is \dfrac{-2}{x^2-1}.

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It’ll be better than the one your group created.

Step-by-step explanation:

This is because the curiosity rover did not descend at Constant rate, so a curve is most preferred than a line.

4 0
3 years ago
Jerome studied math for5/8 of an hour grammar for 1/2 of an hour science for 2/5 of an hour and history for 3/10 of an hour. wha
Verizon [17]
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3 years ago
How can we find different ways to present algebraic expressions without changing their meaning
HACTEHA [7]

Answer:

The algebraic expressions can be presents by words or equivalent expressions, so that its meaning remains same.

Step-by-step explanation:

Different ways to present algebraic expressions without changing their meaning:

1. We can describe the algebraic expressions in words so that its meaning remains same.

For example : 3x+y = Sum of 3 times of an unknown variables x and y.

2. We can describe the algebraic expressions in equivalent expressions, so that its meaning remains same.

For example : 6x+3y = 3(2x+y)

Therefore, the algebraic expressions can be presents by words or equivalent expressions, so that its meaning remains same.

4 0
3 years ago
Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.
kupik [55]
The distance between some point (x,y,z) and the origin is given by

f(x,y,z)=\sqrt{x^2+y^2+z^2}

so this is the function we're trying to minimize. But notice that f(x,y,z) and f(x,y,z)^2 attain their critical points at the same (x,y,z), so we can solve the same problem by minimizing x^2+y^2+z^2 instead.

So let's take the Lagrangian to be

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)

with partial derivatives (set equal to 0)

L_x=2x+a\lambda=0
L_y=2y+b\lambda=0
L_z=2z+c\lambda=0
L_\lambda=ax+by+cz-d=0

Now, notice that

aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0
\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}

and we can use this to solve for x,y,z. We get

x=\dfrac{ad}{a^2+b^2+c^2}
y=\dfrac{bd}{a^2+b^2+c^2}
z=\dfrac{cd}{a^2+b^2+c^2}

At this critical point, we get a minimum distance of

\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}
8 0
3 years ago
Brainliest to the most correct answer!!!
tino4ka555 [31]

Answer:

Multiply the top equation by two because both y value will become opposites and will eliminate.

Please give Brainliest! I never got one! :)

3 0
3 years ago
Read 2 more answers
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