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3 years ago

Caitlyn used a rubber band and geoboard to make the rectangle shown. What is the area of the rectngle

Mathematics

1 answer:

alex41 [277]3 years ago

6 0

Length of the rectangle multiplied by the width so L*W = area

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There are 180 school days in a year. How many days are in 5 years

labwork [276]

There are 180 school days in a year. How many school days are in 5 years?

If we know that there are 180 school days in 1 year, we can multiply 180 by 5 to get our answer of how many school days are in 5 years.

900 school days in 5 years.

5 0

3 years ago

What is the value of 2x^3+7x^2-19x-6 if the value of x is -1

Paul [167]

Answer:

-40

Step-by-step explanation:

Let's input x into all of the slots.

$-2^{2}+-7^{2}+19-6$

Now, let's solve!

$-4-49+19-6$

$-40$

So, the answer is -40!

5 0

3 years ago

R=5(CA−0.3) solve for c

Maurinko [17]

R = 5(CA - 0.3)

<span>Multiply everything in the parenthesis by 5.</span>

R = 5CA - 1.5

Add 1.5 to both sides

1.5 + R = 5CA

Divide 5A on both sides.

C = 1.5 + R / 5A

Hope this helps!

4 0

3 years ago

Read 2 more answers

2 ( 3 x 4) - 10 + 14/2 = 21<br><br> true or false

larisa86 [58]

Answer:

true

Step-by-step explanation:

7 0

3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of

Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

$P(t) =504,943.26 -104,943.26e^{0.693t}$

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

$\implies \frac{dP}{dt} =kP$ ----(1)

where k is constant rate at which population is doubled

solving (1)

$ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}$

$t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\$ ---- (2)

initial population = 400,000

population is doubled every week

⇒P(1)=2P(0)

Using (2)

$P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\$

$e^{k} =2\\k=ln|2|\\$

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

$\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\$ ---(3)

solving (3) by calculating integrating factor

$I.F=e^{\int-k dt}$

Multiplying I.F with all terms of (3)

$e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}$

Integrating w.r.to t

$e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C$

$P(t) =\frac{350,000}{k} +Ce^{kt}\\$

$k=ln|2| =0.693$

$P(t) =504,943.26 + Ce^{0.693t}\\$

at t=0

$P(0) =504,943.26 + Ce^{0.693(0)}$

$400,000 =504,943.26 + C$

$C = -104,943.26$

So, population of mosquitoes in the area at any time t is

$P(t) =504,943.26 -104,943.26e^{0.693t}$

6 0

3 years ago