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2 years ago

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What is the rate of change??

1 answer:

N76 [4]2 years ago

8 0

Answer:

$-2 = m$

Step-by-step explanation:

Starting from the origin, you do $\frac{rise}{run}$ by either moving two blocks <em>north</em> over one block <em>west</em><em> </em>or two blocks <em>south</em> over one block <em>east</em> [<em>west</em> and <em>south</em> are negatives].

OR

$\frac{-y_1 + y_2}{-x_1 + x_2} = m$

[2, −4] and [−2, 4]

$\frac{-4 - 4}{2 + 2} = -\frac{8}{4} = -2$

I am joyous to assist you anytime.

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Find the greatest common factor 32a^3b^2, 44ab^3 and 40a^2b

Ronch [10]

$\32a^3b^2 \ , \ 44ab^3 \ , \ 40a^2b\\
$

$\text {Taking out common factor : } \\ \\
4ab(8a^2b) + 4ab(11b^2) + 4ab(10a)$

$\text { greatest common factor = } 4ab$

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3 years ago

A bobsleds distance-time graph indicates that it traveled 100m in 25 s what is the bobsleds speed?

Brilliant_brown [7]

The speed is 4 meters per second

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3 years ago

Given: ABCD is a ∥-gram, BF ⊥ CD , BE ⊥ AD, Prove: △ABE∼△CBF

drek231 [11]

Answer:

Hence proved △ABE∼△CBF.

Step-by-step explanation:

Given,

ABCD is a parallelogram.

BF ⊥ CD and

BE ⊥ AD

To Prove : △ABE∼△CBF

We have drawn the diagram for your reference.

Proof:

Since ABCD is a parallelogram,

So according to the property of parallelogram opposite angles are equal in measure.

$\therefore m\angle A = m\angle B$ ⇒1

And given that BF ⊥ CD and BE ⊥ AD.

So we can say that;

$m\angle F=m\angle E=90\°$ ⇒2

Now In △ABE and △CBF

∠A = ∠C (from 1)

∠E = ∠F (from 2)

So by A.A. similarity postulate;

△ABE∼△CBF

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

2 years ago

In need of help pls. Try and answer asap.

Alina [70]

Answer:

Step-by-step explanation:

∠1 and 85 are supplementary

m∠1 = 180 - 85 = 95°

4 0

2 years ago