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dedylja [7]
2 years ago
8

The equation x2 – 1x – 90 = 0 has solutions {a, b}. What is a + b

Mathematics
2 answers:
Andrews [41]2 years ago
8 0
X^2 - x - 90 =0
This equation is in standard form ax^2+ bx + c
The sum of the solutions is -b/a (a and b are the coefficients from the equation above, not the solutions to the equation)
The answer is 1/1 = 1
REY [17]2 years ago
3 0

Answer:

Step-by-step explanation:

Alright, let get started.

The given equation is : x^2-1x-90=0

This is a quadratic equation.

The standard form of a quadratic equation is : AX^2+BX+C=0

Comparing the given equation with standard one, the values of A ,B and C are :

A is 1

B is -1

C is -90

Since a and b are the roots of the equation, so the formula for sum of roots a+b will be = \frac{-B}{A}

a+b=-\frac{-1}{1} =1

hence a+b will be 1   :   Answer

Hope it will help :)

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The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours
ella [17]

Answer:

137

Step-by-step explanation:

Use the model A = Pe^(kt), where P is the initial number and k is the growth constant.  Then, in case 1,

300 = Pe^(k*3)

and in case 2,

4000 = Pe^(k*10).  We need to find P and e^k.

300 = Pe^(k*3) can be rewritten as (300/P) = [e^k]^3, which in turn may be solved for e^k:

∛(300/P) = e^k

Now we go back to 4000 = Pe^(k*10) and rewrite it as (4000/P) = [e^k]^10.  Substituting ∛(300/P) for e^k, we obtain:              

(4000/P) =  (300/P )^(10/3)

which must now be solved for P.  Raising both sides by the power of 3, we get:

(4000/P)^3 = (300/P)^10.  Therefore,

4000^3       300^10

------------- = -------------

   P^3             P^10

which reduces to:

4000^3       300^10

------------- = -------------

       1             P^7

or

        1         P^7

------------  =  -----------

4000^3       300^10

or:

           300^10

P^7 = ---------------- = 9.226*10^13  = 92.26*10^12

           4000^3

Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or

                                                                        P = 137.26

The initial number of bacteria was 137.

7 0
2 years ago
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