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3 years ago

11

(18x 5 + 6x 4 - 12x 3) ÷ 6x 2

2 answers:

Gnom [1K]3 years ago

8 0

Your answer would be 26.
if u want reasoning just ask!

astraxan [27]3 years ago

5 0

If I am reading this correctly but I might not because those could be x’s not times. If you follow PEMDAS(please excuse my dear aunt sally helps you remember) you start with parentheses then move to exponents then do you multiplication and Divison and addition and subtraction. So (18 x 5 + 6 x 4 - 12 x 3) you start there 18 x 5 is 90, 6 x 4 is 24, 12 x 3 is 36 so then add 90 +24 which is 114 and then minus 36 from 114. Which is 78. After all that go to 6 x 2 which is 12 from there you would have 78 divided by 12 which equals 6.5. Hopefully that helps but since I could tell if the they were x’s or times.

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Find the area of a circle with a diameter of 8 in. Rounded to the nearest tenth. Use 3.14 for π.

rodikova [14]

Answer:

50.2 in^2

Step-by-step explanation:

Using the Formula

A=πr2

d=2r

Solving for A

A= 1/4 π d^2 = 1/4*π*8^2 = 50.26548

=50.2in^2

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

At 5:00 a.m., the temperature outside was 66° F. By noon, the temperature had risen to 87° F. Which equation models the temperat

likoan [24]

<span>If so, 87 - 66 = 21 / (12-5) = 3 degrees Fahrenheit per hour.</span>

4 0

3 years ago

Read 2 more answers

Farrah's annual salary is £27,750. how much does she get paid per week?

VashaNatasha [74]

Hello human

Answer:

532.19 per week

Explanation:

You divide 27750 by 52.143 Because there are 52.143 weeks per year.

Thank you and gn ❤️✌

7 0

3 years ago

Okay can someone help me with 19 the answer is height = 5 but someone please explain the process to answer

sertanlavr [38]

Answer:

hope it helps you see the attachment for further information

8 0

3 years ago