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RideAnS [48]
3 years ago
13

What are the real zeros of f(x) = x^3 +2x^2-5x-6

Mathematics
1 answer:
netineya [11]3 years ago
8 0

First of all Horner
First we try with -3
1. 2. -5. -6.
-3. 3. 6
1. -1. -2. 0
To verify: -3^3+2*-3^2+15-6=0

Now we have this equation
(X+3)*(x^2 - x -2)=0

I dont know If you know sum and product but that is the easiest way so sum= 1 product = 2 so 2 and -1 are the answers.
You can do this too by this way : b^2-4ac= D
X= (-b(+or-)d^1/2)/2a

So all the zeros
2, -1 and -3
If you want more explanation, send me a comment x
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2 years ago
75c- 300 - 35 c =25c + 15c + 200
GarryVolchara [31]

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4 0
2 years ago
A certain quantity decays exponentially over time. The initial quantity at t = 0 is 25,000. The quantity decays at a rate of 5%.
7nadin3 [17]
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You end up with an equation looking like
25000*0.95^t
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0.95 is being raised to the power of t to determine how many times it has decayed
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3 0
3 years ago
Read 2 more answers
Suppose f(x) = x^2 and
SpyIntel [72]

Given:

The two functions are:

f(x)=x^2

g(x)=\left(\dfrac{1}{3}x\right)^2

To find:

The statement that best compares the graph of g(x) with the graph of f(x).

Solution:

The horizontal stretch is defined as:

g(x)=f(kx)            ...(i)

If 0, the function f(x) is horizontally stretched by factor \dfrac{1}{k}.

If k>1, the function f(x) is horizontally compressed by factor \dfrac{1}{k}.

We have,

f(x)=x^2

g(x)=\left(\dfrac{1}{3}x\right)^2

Using these functions, we get

g(x)=f\left(\dfrac{1}{3}x\right)         ...(ii)

On comparing (i) and (ii), we get

k=\dfrac{1}{3}

Since 0, the function f(x) is horizontally stretched by factor \dfrac{1}{\frac{1}{3}}=3.

Hence, the correct option is D.

5 0
2 years ago
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