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Vanyuwa [196]
3 years ago
15

Graph the function represented by the equation y=3×2-6×-9

Mathematics
1 answer:
sasho [114]3 years ago
7 0
The points in the graph are (-1,0) and (3,0). I solved the equation using the Factoring method in solving quadratic equations. As I solved for finding x= -1 and 3, I substituted both to the equation in order to find y. 
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ifhofgoefouegf

Step-by-step explanation:

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Revolving credit lines open account credit offered by banks or other financial institutions.
You should have <span>2 to 6 </span>revolving credit lines.
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N is a rational number.m is an irrational number.n2 cannot be an irrational number? why or why not
Korolek [52]

If n is rational, it means that

n\in\mathbb{Q}\Rightarrow n=\frac{p}{q},\quad p\in\mathbb{Z},\quad q\in\mathbb{Z}^*

Therefore when we do n² we can write it as

n^2=\frac{p^2}{q^2},\quad p\in\mathbb{Z},\quad q\in\mathbb{Z}^*

Remember that the product of two integer numbers is also an integer, therefore we can guarantee that

p^2\in\mathbb{Z}\text{ and }q^2\in\mathbb{Z}^*

Then we can confirm that n² is the quotient of two integers and the denominator is not zero, therefore, n² is always rational, it cannot be an irrational number

8 0
1 year ago
Write a linear equation for a line that passes through point(2, 3) with a slope of 1/2
aalyn [17]

Answer:

y = 1/2x + 2

Step-by-step explanation:

y = 1/2x + b

3 = 1/2(2) + b

3 = 1 + b

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8 0
3 years ago
The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon
Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by x=3.24t-4.20t^2+1.07t^3 where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

x(t)=3.24t-4.20t^2+1.07t^3

x(1)=3.24(1)-4.20(1)^2+1.07(1)^3

x(1)=3.24-4.20+1.07

x(1)=0.11

b) At t= 2 s

x(t)=3.24t-4.20t^2+1.07t^3

x(2)=3.24(2)-4.20(2)^2+1.07(2)^3

x(2)=6.48-16.8+8.56

x(2)=-1.76

c) At t= 3 s

x(t)=3.24t-4.20t^2+1.07t^3

x(3)=3.24(3)-4.20(3)^2+1.07(3)^3

x(3)=9.72-37.8+28.89

x(3)=0.81

d) At t= 4 s

x(t)=3.24t-4.20t^2+1.07t^3

x(4)=3.24(4)-4.20(4)^2+1.07(4)^3

x(4)=12.96-67.2+68.48

x(4)=14.24

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

\triangle x=x(4)-x(0)

\triangle x=14.24-0

\triangle x=14.24

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity  is given by,

\triangle x=x(4)-x(2)

\triangle x=14.24-(-1.76)

\triangle x=14.24+1.76

\triangle x=16

4 0
3 years ago
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