All categories

3 years ago

Help please!!! can you show me how to do this??

Mathematics

1 answer:

IrinaK [193]3 years ago

5 0

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

You might be interested in

ANSWER ASAP THANKS ...............

mylen [45]

Answer:

250

Step-by-step explanation:

5*5=25

25*10=250

+$++$$+$

6 0

3 years ago

Can someone please explain how to do these?

Pani-rosa [81]

Answer:

First question answer: The limit is 69

Second question answer: The limit is 5

Step-by-step explanation:

For the first limit, plug in $x=8$ in the expression $(9x-3)$ , that's the answer for linear equations and limits.

So we have:

$9x-3\\9(8)-3\\72-3\\69$

The answer is 69

For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value $x=1$ into the simplified expression to get the correct answer. Shown below:

$\frac{x^2+8x-9}{x^2-1}\\\frac{(x+9)(x-1)}{(x-1)(x+1)}\\\frac{x+9}{x+1}$

Now putting 1 in $x$ gives us the limit:

$\frac{x+9}{x+1}\\\frac{1+9}{1+1}=\frac{10}{2}=5$

So the answer is 5

3 0

3 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

Slope:

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

Slope:

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$

So the diagonals measure the same, therefore this is a rectangle.

5 0

3 years ago

Please help and show work !!!!

maria [59]

Answer and work is in the picture !

3 0

3 years ago

Read 2 more answers

Jerry and Katja are running in a race. Jerry runs really fast for a while. Then, he stops to catch his breath. After he has rest

dimaraw [331]

Katja would be the straight line and joels line would show sections that stop going up it would intersect at some point because katja is constantly moving while joel takes breaks in-between

Hope this helps!! :))

7 0

4 years ago

Read 2 more answers