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Paraphin [41]
3 years ago
5

Write the polynomial equation for -1/5,-1/2,-4​

Mathematics
1 answer:
ycow [4]3 years ago
5 0

Answer:

(5x+1)(2x+1)(x+4)

Step-by-step explanation:

The factored from of a polynomial can be found from the zeros or x-intercepts of the graph.

The x-intercepts here are x= -1/5, x=-1/2 and x= -4. You can find the factors by finding the expression which solves for each value.

Since x = -1/5 then the factor is 5x+1.

Since x = -1/2 then the factor is 2x+1.

Since x = -4 then the factor is x+4.

So the factored form is (5x+1)(2x+1)(x+4).

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Ghella [55]
You can figure this out by turning \frac{3}{5} into a decimal you do that by dividing the numerator by the denominator so 3 divided by 5 equals 0.6 so now which is grater 0.7 or 0.6 
0.7 is greater than \frac{3}{5}
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3 years ago
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Show me how to do it please
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Set the two "y='s" to each other and then solve for x.
To start it off you would do this: 6x-2=-6x+6
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3 years ago
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Which is an equation for the line with a slope of 1/3, and passes through<br> the point (–2, 1)?
worty [1.4K]

Answer:

y = (1/3)x +5/3

Step-by-step explanation:

The general equation of a line is

y = mx+b

where m is the slope and b the y-intercept

Since the slope is 1/3 we have

y = (1/3)x +b

since the line passes through (x,y) = (-2,1) this point satisfies the equation

1 = (1/3)(-2) + b ===> b = 1+2/3 ===> b = 5/3

and the equation of the line is

y = (1/3)x +5/3

8 0
3 years ago
3 sides of the triangle are distinct prime numbers. What is the smallest possible perimeter of the triangle?
Nitella [24]

Answer:

the smallest possible perimeter is 6

Step-by-step explanation:

since the smallest possible prime number is 2 it would make since if each side is 2 which makes the smallest perimeter possible. i really hope this helped :)

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Form quadratic equation with roots (alpha² + 1/alpha²) and (beta²+1/beta²)​
sashaice [31]

Answer:

What equation? Here’s a big tip for you, equations contain an ‘equals sign’ (‘=’) and something that they are equal to. I’m assuming your equation is:

ax2+bx+c=0  

I’m going to use  p  and  q  for the roots of the equation as they are easier to type than  α  and  β . Thus we have:

a(x−p)(z−q)=0⇒ax2−a(p+q)x+apq=0  

Equating coefficients with our initial equation:

x1 :  b=−a(p+q)⇒p+q=−ba  

x0 :  c=apq⇒pq=ca  

 

Now  p2+q2=p2+2pq+q2−2pq=(p+q)2−2pq  

=(−ba)2−2ca  

=b2a2−2ca  

=b2−2aca2

3 0
3 years ago
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