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2 years ago

Factor the GCF: −12x4y − 9x3y2 + 3x2y3 (5 points)

Mathematics

1 answer:

garri49 [273]2 years ago

8 0

$-12x4y - 9x3y^2 + 3x^2y^3 = 3xy(-16 -9y + xy^2)$

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What is the product of 2/3 and 4/5?

Paul [167]

Well, product means multiply so we need to multiply these fractions.

2 4
--- x --- =
3 5

now, when you are multipliying fractions you multiply the top and the bottom with the other one.
so 2 times 4 and 3 times 5
and you get
8
--
15

since this fraction can not be simplifyed this is the final answer 8/15

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3 years ago

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How do you factor 45x^2+295x-140

Vaselesa [24]

I could explain it, but you have to ask me, other than that the answer is 5(x+7)(9x-4).

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2 years ago

PLZ HELP ME HAS SOON HAS POSSIBLE?????

olganol [36]

Answer:

- 7/2

Step-by-step explanation:

- sqrt(49/4)

We know that sqrt(a/b) = sqrt(a)/ sqrt(b)

- sqrt(49)/sqrt(4)

- 7/2

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2 years ago

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Solve by graphing. x2 2x - 8 = 0 a. –2, 4 b. 2, –4 c. 2, –4 d. –2, 4

Luba_88 [7]

Well, to solve, garph the equation y=x^2+2x-8 and y=0
see where they intersect

anyway the easier way is to factor but anyway
0=(x-4)(x+2)

anyway

solution is x=4 and -2

A and D are answers (they are same)

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2 years ago

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

ollegr [7]

Answer:

The population of mosquitoes in the area at any time t is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of k as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of C as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute C in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

6 0

2 years ago