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4 years ago

9

What are the x-intercepts of the graphs of the function shown below y=x^2+4x-32?

1 answer:

stellarik [79]4 years ago

5 0

They are x = 8 and x = -4

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Please help! i literally suck at geometry

sdas [7]

Don’t say like that!Think you can and you will achieve! :) :)

8 0

3 years ago

4.6.3 Test (CST): Linear Equations

inn [45]

The slope of the green line if the lines are perpendicular is -1/4

<h3>Perpendicular lines</h3>

For two lines two be perpendicular, the product of their slope must be -1. Let the slope of the red and green line be m1 and m2.

Given the following

Slope of red line = 4

According the definition

4m2 = -1

m2 = -1/4

Hence the slope of the green line if the lines are perpendicular is -1/4

Learn more on perpendicular lines here: brainly.com/question/1202004

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5 0

2 years ago

Please help me :( with this

hodyreva [135]

Answer:

21

Step-by-step explanation:

Similar triangles. MNL is just ABC but 3/4 the size.

x = 8*3/4 = 6

perimeter woudl be 6+6+9 = 21

7 0

3 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

3 years ago

Question 8 is the one I’m having trouble on it

salantis [7]

Just multiply 30 times 2,000 and also multiply 30 times 700. The reason you will multiply it by 30 is because in 1 month there is 30 days. 30 x 2,000= 60000 30 x 700 = 21000 I’m only 9 btw so sorry if I didn’t give a lot of detail lol

4 0

3 years ago