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vodomira [7]
2 years ago
13

Write an equation of the line in slope-intercept form (-3,3) (0,-1)

Mathematics
1 answer:
Vesnalui [34]2 years ago
3 0
Y= 4/3x -1 use y2-y1 over x2-x1

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PLEASE IM BEGGING! HELP!
user100 [1]

Answer:

Part A: after 9 days the radius of the algae would be 12.81mm so the domain is 9. you would plot the domain at (0,9)

Part B: the y-intercept (the 9) represents the amount of algae the experiment started off with.

Part C: the rate of change is 0.

Step-by-step explanation:

4 0
3 years ago
Find the rate of change for this linear function Y equals MX plus be
Crazy boy [7]
Y=mx+b is an equation 
8 0
4 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
Read 2 more answers
Solve the system of two linear inequalities graphically.
Sladkaya [172]

Answer:

See attachment

Step-by-step explanation:

Isolate y in the first inequality:

2x + 4y < -16\\4y < - 16 -2 x\\y < -4 -\frac{1}{2} x\\y < -\frac{x+8}{2} \\

Now, with both x and y inequalities found, graph it.

7 0
2 years ago
15-12 + 4(3) =<br> 15-12-12<br> 15-1<br> 14
MissTica

Answer:

9

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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