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levacccp [35]
3 years ago
14

I need to write the equation in slope intercept form that passes through the points

Mathematics
1 answer:
VMariaS [17]3 years ago
4 0
(x1, y1) (x2, y2)
(-4, 5)    (-1, 1)
x2-x1 
-1 - -4
-1+4
   3
y2-y1
1- -4
1+4
  5
 
answer : (3,5)
I believe this is right, hope this helps!


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This is the answer hope it helps you

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Give the largest interval over which the general solution is defined. (Think about the implications of any singular points. Ente
Leto [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   No singular point due to the exponent in the solution

    The interval is   -\infty

b

   NONE

Step-by-step explanation:

From the question we are told that

     \frac{dy}{dx} =  9y

The generally solution is mathematically represented as

         \frac{dy }{dx}  =  9y

=>       \frac{dy}{y}  =  9dx

integrating both sides  

         \int\limits  {\frac{ dy}{y} } \,  = \int\limits  {9} \, dx

  =>   lny = 9x + c

 =>   y =  e^{9x +c }

 =>    y =  e^{9x} e^{c}

Here e^c  =  C

=>     y = C  e^{9x}

From the above equation we see that the domain for x has no singular point the interval is

       -\infty

Also there is no transient term in the general solution obtained because as  x \to \infty there no case where y \to 0

7 0
3 years ago
What is the equation of this line in slope-intercept form?
Elis [28]

Answer:

y=-1/2x+2

Step-by-step explanation:

for x-intercept

put y=0

0= -1/2x+2

1/2x=2

x=4

(4,0)

for y-intercept

put x=0

y= 2

(0,2)

7 0
3 years ago
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Which is the graph of the function: f(x) = x2 - 4x + 3​
Nina [5.8K]

Answer:

Option 1

Step-by-step explanation:

To find the graph of the quadratic function, we find it's zeros.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}

x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}

\Delta = b^{2} - 4ac

f(x) = x² - 4x + 3​

This means that a = 1, b = -4, c = 3

So

\Delta = b^{2} - 4ac = (-4)^2 - 4(1)(3) = 16 - 12 = 4

x_{1} = \frac{-(-4) + \sqrt{4}}{2} = 3

x_{2} = \frac{-(-4) - \sqrt{4}}{2} = 1

Zeros at x = 1 and x = 3, that is, it crosses the x-axis at this values, so the graph is given by option 1.

4 0
3 years ago
What’s the answer plz answer
goldfiish [28.3K]
The answer is 3 because the y intercept is 2 and y/x would be up one over two
6 0
3 years ago
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