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Lesechka [4]
3 years ago
10

for the equation 2x(3x+5)+3(3x+5)=ax^2+bx+c , if this equation is true for all values of x,what is the value of b if a,b,and c a

re constants
Mathematics
1 answer:
guapka [62]3 years ago
6 0

Answer:

b=19

Step-by-step explanation:

2x(3x+5)+3(3x+5)=ax2 +bx+c

by expansion,

6x2 +10x +9x+15=6x2 +19x +15

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Step-by-step explanation:

5 0
3 years ago
I need help on question 2 i need the simplified answer and the restrictions
Anastaziya [24]

Answer:

x cannot be 5 or -3.

\frac{x+5}{x+3}

Step-by-step explanation:

The restrictions for a fraction is that the bottom cannot be 0.

So if we find when the bottom is 0 we have found the values that x cannot be.

Let's solve x^2-2x-15=0.

Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.

Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.

So the factored form of the equation is:

(x-5)(x+3)=0

This means either x-5=0 or x+3=0.

We do have to solve both.

x-5=0 can be solved by adding 5 on both sides.

x-5+5=0+5

x+0=0+5

x=5

x+3=0 can be solved by subtracting 3 on both sides.

x+3-3=0-3

x+0=0-3

x=-3

So x can be any number except x=5 or x=-3.

We already factored the bottom as (x-5)(x+3).

The top is a difference of squares, x^2-a^2,

which can be factored as (x-a)(x+a).

So the top factors as (x-5)(x+5).

The fraction can then be written as:

\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}

This can further simplified assuming x is not 5 we can write it as \frac{x+5}{x+3}.  I canceled the common factor of (x-5).

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3 years ago
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Yuliya22 [10]

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The explanation is in the picture below if needed

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2 years ago
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trasher [3.6K]

-40 i think.............. im not positive but i did my best

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3 years ago
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