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3 years ago

How do u multiply x^2-4 over x^2-1 times x+1 over x^2+2x

2 answers:

6 0

$\frac{x ^ 2-4 }{x ^ 2-1}*\frac{ x + 1 }{x ^ 2 +2 x}=\frac{x ^ 2-2^2 }{x ^ 2-1}*\frac{ x + 1 }{x(x +2 )} =\\ \\ \\ =\frac{(x -2)\not(x+2)^1 }{(x -1)\not(x+1)^1}*\frac{ \not(x + 1)^1 }{x\not(x +2)^1 }= \frac{ x-2 }{x(x-1)} \\ \\ \\ \\a^2-b^2 =(a-b)(a+b)$

mixer [17]3 years ago

5 0

(X2-4)/(x2-1). * (x+1)/(x2+2x)
[(x-2)(x+2)]/[(x-1)(x+1)]. * (x+1)[x(x+2)]
(x-2)/(x-1) * 1/x
(x-2)/(x2-x)

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Tan^2 70° -sec^2 70°

kipiarov [429]

Answer:

-1

Step-by-step explanation:

We need to find the value of tan² 70° -sec² 70°.

We know that,

$\sec^2\theta-\tan^2\theta=1$ ...(1)

We can write the given expression as :

tan² 70° -sec² 70° = -(-tan² 70° +sec² 70°)

=-(sec² 70°-tan² 70°)

Using identity (1)

=-(1)

= -1

Hence, the value of the given expression is -1.

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2 years ago

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stiv31 [10]

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2 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint: Endpoint - (1, 3) Midpoint - (0, 0)

Kobotan [32]

Answer:

The other endpoint of the line segment = (-1, -3)

Step-by-step explanation:

A segment has:

Endpoint: (1, 3)

Midpoint: (0, 0)

Let (x, y) be the other endpoint of the line segment

$\frac{\left(1\:+\:x\right)}{2}=0$

$\mathrm{Multiply\:both\:sides\:by\:}2$

$\frac{2\left(1+x\right)}{2}=0\cdot \:2$

$1+x=0$

$x=-1$

$\frac{\left(3+y\right)}{2}=0$

$\mathrm{Multiply\:both\:sides\:by\:}2$

$\frac{2\left(3+y\right)}{2}=0\cdot \:2$

$3+y=0$

$y=-3$

Therefore, the other endpoint of the line segment = (-1, -3)

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3 years ago

There are 9 marbles representing 3 different colors. Write a problem where 2 marbles are selected at random without replcement a

Nata [24]

There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.

Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).

In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.

Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.

Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.

The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?

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3 years ago

22) John ordered a large pizza for $19.50. How much

siniylev [52]

i believe its $21.84 because 12% would be $2.34

Step-by-step explanation:

7 0

3 years ago

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