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3 years ago

-8c - 6 less than or equal to 10

Mathematics

1 answer:

Step2247 [10]3 years ago

3 0

-8c-6<10
-8c<16
C>-2

And there should be a line under the “>”. It flips because you had to decide it by a negative number.

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If h= 1/2mk ,express m in terms of h and k

Sergeeva-Olga [200]

we are given

$h=\frac{1}{2}mk$

Since, we have to solve for m

so, we will isolate m on anyone side

step-1: Multiply both sides by 2

$2*h=2*\frac{1}{2}mk$

$2h=mk$

step-2: Divide both sides by k

$\frac{2h}{k} =\frac{mk}{k}$

$\frac{2h}{k} =m$

so, we get

$m=\frac{2h}{k}$ ..............Answer

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4 years ago

7.7 times 10 to the 12th power

podryga [215]

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3 years ago

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What is the point of maximum growth:

salantis [7]

f(h(x))= 2x -21

Step-by-step explanation:

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

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4 years ago

There were 409 students in school. After some of them graduated, there were 172 students left. How many students graduated ?

nataly862011 [7]

Answer:

237 students graduated

Step-by-step explanation:

Subtract 172 from 409, obtaining 237.

237 students graduated

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3 years ago

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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

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

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3 years ago