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ki77a [65]
3 years ago
11

Multiply

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
6 0

Answer:

The product of given two fractions is

\frac{4(x+2)}{(x-1)(x+3)}

Therefore \frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{4(x+2)}{(x-1)(x+3)}

Step-by-step explanation:

Given expression is  

\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}

To find the product of two given fractions as below :

\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{2(5)}{x(x-1)}\times \frac{2x(x+2)}{5(x+3)}

=\frac{10}{x(x-1)}\times \frac{2x(x+2)}{5(x+3)}

=\frac{20x(x+2)}{5x(x-1)(x+3)}

=\frac{20x^2+40x}{(5x^2-5x)(x+3)} (multiply each term in the factor to each term in the another factor )

=\frac{20x^2+40x}{5x^3+15x^2-5x^2-15x}  ( adding the like terms )

=\frac{20x^2+40x}{5x^3+10x^2-15x}

=\frac{5x(4x+8)}{5x(x^2+2x-3)}

=\frac{4(x+2)}{(x-1)(x+3)}

Therefore \frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{4(x+2)}{(x-1)(x+3)}

Therefore the product of given two fractions is \frac{4(x+2)}{(x-1)(x+3)}

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3 years ago
Three lights all go on one changes every 5 sec one every 6 sec one every 8 sec when will they all go on together
inysia [295]
In order to calculate, you need to find their LCM
LCM of 5,6 and 8 is equal to 120

So, That time will be 120 sec = 2 min.
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Hope this helps!
3 0
3 years ago
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Andrej [43]
Your answer for this is 17 hope i helped
8 0
3 years ago
Read 2 more answers
Choose the graph that represents the following system of inequalities: y ≤ −3x + 1 y ≤ 1 over 2x + 3 In each graph, the area for
Vanyuwa [196]

The graph that represents the inequality has been shown in the attachment.

<h3>How to solve for the graph</h3>

We have these equations

y ≤ −3x + 1

y ≤ x + 3

We remove the inequality sign from both of these equations

y = −3x + 1

y = x + 3

−3x + 1  = x + 3

such that

x = -0.5

we use this value for x in any of the equations

x + 3 = -0.5 + 3

= 2.5

the point of intersection is at 2.5, -0.5

we test for the origin. 0,0

3x + 1

= 3*0 + 1

= 1

for  x + 3

0+3 = 3

This is 0≤1 and 0≤3

Hence the graph should be shaded to the origin.

Read more on a graph here: brainly.com/question/14030149

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7 0
2 years ago
A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ
Naya [18.7K]

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = 2.2\times 7.4=16.28

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = 2.2\times -6.2=-13.64

Thus change in moment is: 16.28-(-13.64)=29.92

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

\frac{29.92}{0.25}=119.68

Hence, the constant force exerted on the ball by the wall is 119.68 N.

6 0
3 years ago
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