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3 years ago

PLZ HELPP MARK BRAINLIEST 20 POINTS

Mathematics

1 answer:

morpeh [17]3 years ago

3 0

Answer:

This is incorrect as 1/8 plus 3/5 is 29/40

Step-by-step explanation:

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Compare these fractions by filling in the blank with the appropriate sign. 35 _____ 56

BartSMP [9]

Answer:

35 < 56.

56 is greater than 35.

I hope this helped at all.

5 0

3 years ago

What transformation can be used to transform figure ABCDEF to A"B"C"D"E"F

morpeh [17]

Answer:

ABCDEFG

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP ; I need help on this ? I am confused .

zubka84 [21]

256,000-2.56*100,000
Since there are 6 digits in 256,000, it is 2.56*100,000 becuase 100,000 also has 6 digits.
2.56*100,000=2.56*10^5
Since there are 5 zeroes in 100,000, it is 10^5.
256,000=2.56*100,000=2.56*10^5
First box: 100,000
Second box: 5

6 0

3 years ago

Read 2 more answers

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

I NEED HELP PLEASE I WILL MARK BRAINLIST The question is in the screenshot

Ivan

The answer is unreasonable because the value of x is between 2 and 3.The error is : she did not introducing logs in solving the equation.

Step-by-step explanation:

The question is $7^x=95$

Here answers is unreasonable because, she can test with powers of seven like;

7²=49

7³=343

So this means the value of x lies between 2 and 3, so here answer 11.71 is unreasonable

b) In this case, to solve the equation, introduce logs both sides

$7^x=95$

$log7^x=log95\\\\\\xlog7=log95\\\\\\x=\frac{log95}{log7} \\\\\\x=2.340$

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Indices :brainly.com/question/1660321

Keywords :Equation, error

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4 0

3 years ago