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3 years ago

What is the difference A B C D

Mathematics

1 answer:

aksik [14]3 years ago

4 0

Answer:

D. $\frac{x-1}{x^2+3x+2}}$

Step-by-step explanation:

The given expression is;

$\frac{x}{x^2+3x+2}-\frac{1}{(x+2)(x+1)}$

This is the same as;

$\frac{x}{(x+2)(x+1)}-\frac{1}{(x+2)(x+1)}$

The denominators are the same.

Subtract the numerators and write one denominator;

$\frac{x-1}{(x+2)(x+1)}}$

Rewrite to get;

$\frac{x-1}{x^2+3x+2}}$

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Evaluate each expression if a=-1, b=4,x=-2 and y=3 2(6a-2b)-8x

Andreyy89

Answer:

-12

Step-by-step explanation:

2(6a-2b)-8(-2)

=2[6(-1)-2(4)]+16

=2[-6-8]+16

=2[-14]+16

=-28+16

=-12

4 0

3 years ago

Decide whether the statement is true or false. If it is false, explain why. The union of the solution sets of 4xplus5equals13,

postnew [5]

The given statement is false because it isn't an empty set!

Step-by-step explanation:

We have following sets of inequalities:

$4x+5=13\\4x+5>13\\4x+5$

From $4x+5=13$ we get ,

$4x = 8 \\x=2$

Therefore solution set is x=2.

Now, for $4x+5>13$ we get ,

$4x+5>13 \\4x>8\\x>2$

Therefore solution set is x>2.

For $4x+5$ we get ,

$4x+5$

Therefore solution set is x<2.

Now, the union of x=2, x<2 & x>2 is -∞<x<∞. i.e. all possible values of x. And so above statement is false because it isn't an empty set!

4 0

2 years ago

Perform the multiplication FOR THIRTY POINTS! (x^2)/(3) multiplied by (x^2-x-6)/(x^2-6x+9)

Oxana [17]

Answer:

(x^3+2x^2)/(3x-9)

Step-by-step explanation:

3 0

3 years ago

Please help ! find the value of x.

patriot [66]

Answer:

a straight line i think is always 180 so i think the answer is A. 81

Step-by-step explanation:

5 0

3 years ago

(a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve

Dafna1 [17]

Answer:

(a) y = 9/4x -1/4

(b) (-2, -2), (-2, 2)

Step-by-step explanation:

(a) The derivative can be found from ...

2y·y' = 3x² +6x

At (x, y) = (1, 2), this is ...

4y' = 9

y' = 9/4

so the equation of the tangent is ...

y = (9/4)(x -1) +2

y = (9/4)x -1/4

(b) At y' = 0, we have ...

0 = 3x² +6x = 3x(x +2)

This says y' = 0 at x=0 or x=-2. (x = 0 is an extraneous solution.)

At x = -2, we have ...

y² = (-2)³ +3(-2)² = -8+12 = 4

y = ±2

So, the horizontal tangent points are (x, y) = (-2, ±2).

5 0

3 years ago