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3 years ago

Which is a factor of 24x2 − 44x + 12

Mathematics

2 answers:

borishaifa [10]3 years ago

5 0

Answer:

(2x-3) and (3x-1)

Step-by-step explanation:

The given expression is 24x² - 44x + 12

Now we have to factorize it

24x² - 44x + 12 = 4(6x² - 11x + 3)

= 4[3x(2x-3)-1(2x-3)]

= 4[(2x-3)(3x-1)]

Therefore, (2x-3) and (3x-1) are the factors of the given expression.

lutik1710 [3]3 years ago

3 0

4(2x-3)(3x-1) this is the answer

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Find x of this shape.

bagirrra123 [75]

(n - 2)180 = sum of angles (six sides)

(6 - 2)180 = y

(4) 180 = y

720= y

115+110+x+25+x-30+x+x = 720

4x + 220 = 720

4x = 720 - 220

4x = 500

x = 500/4

x = 125

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3 years ago

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Question 4-6. Earth is 93 million miles from the Sun. Pluto is 4.67 billion miles from the Earth.Calculate the ratio of the dist

luda_lava [24]

Answer:

PE/SE = 5.02×10¹

Step-by-step explanation:

Sun-Earth (SE) = 93×10⁶ miles

Pluto-Earth (PE) = 4.67×10⁹ miles

so, PE/SE = (4.67×10⁹) / (93×10⁶) = (4.67×10³)/93 =

= 50.215053763440860215... =

= 5.02×10¹ / 1 or simply 5.02×10¹

in other words, Pluto is a little bit more than 50 times farther away from Earth, than Earth is from the Sun.

6 0

3 years ago

I don't know what the answer is to divide 180 in a ratio of 4:5:9

Karolina [17]

Hello there,

First you need to add together the numbers.
4+5+9 = 18
Divide the total by this,
180 / 18 = 10
Multiply each number by 10.
40:50:90.

Hope This Helps You!
Good Luck Studying :)

4 0

4 years ago

Read 2 more answers

The number line represents Allie’s scores in the first two rounds of a card game.

olganol [36]

Answer:

Allie gains 4, she loses 4, she has 0 points.

Step-by-step explanation:

The arrow shows it. Round 1 goes 0 to 4. Round 2 goes 4 to 0.

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