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Natasha2012 [34]

2 years ago

15

Find roots of 4x^2-2x+9

1 answer:

Korolek [52]2 years ago

4 0

discriminant= b^2-4ac= (-2)^2-4(4)(9)= -140<0

since discriminant <0, graph is always positive, x has no real roots

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Solve the system y = -x + 7 and y = -0.5(x - 3)^2 + 8.

AlladinOne [14]

Answer:

(1,6) & (7,0)

Step-by-step explanation:

y = -x + 7

y = -0.5(x - 3)² + 8

To solve the system, solve these two equations simultaneously

-x + 7 = -0.5(x - 3)² + 8

-x + 7 = -0.5(x² - 6x + 9) + 8

-x + 7 = -0.5x² + 3x - 4.5 + 8

0.5x² - 4x + 3.5 = 0

x² - 8x + 7 = 0

x² - 7x - x + 7 = 0

x(x - 7) - (x - 7) = 0

(x - 1)(x - 7) = 0

x = 1, 7

y = -1 + 7 = 6

y = -7 + 7 = 0

(1,6) (7,0)

Since the system has two distinct solutions, the line and the curve meet at two distinct poibts9: (1,6) & (7,0)

8 0

2 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

2 years ago

Plz help plz asap lol HAHAHA plz

NISA [10]

Answer:

Mario ate 27.27% of her cookies

8 0

3 years ago

Help please!<br> <br> Lines a and b are parallel.<br> <br> What is the measure of angle m?

AlexFokin [52]

So the transversal (line that crosses lines a and b) is 180 since its a straight line. given the angle degree 125, you can do 180-125 to get angle m.

8 0

2 years ago

3x+y=1/3<br> 2x-3y=8/3<br><br> Solve the system of linear equation using any method.

krek1111 [17]

3x + 1y = ¹/₃   ⇒ 9x + 3y = 1
2x - 3y = 2²/₃ ⇒ 2x - 3y = 2²/₃
   11x = 3²/₃
      11    11
         x = ¹/₃
   3x + y = ¹/₃
         3(¹/₃) + y = ¹/₃
   1 + y = ¹/₃
   - 1 - 1
   y = ⁻²/₃
      (x, y) = (¹/₃, ⁻²/₃)

4 0

3 years ago