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garri49 [273]
3 years ago
9

36d^3+24 solve please.

Mathematics
2 answers:
Oduvanchick [21]3 years ago
5 0
The answer is:

12 (3d^3 + 2)

pls mark as brainlist!!
Sergeeva-Olga [200]3 years ago
3 0

Solution, 36d^3+24:\quad 12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Steps:

36d^3+24

\mathrm{Factor\:out\:common\:term\:}12,\\12\left(3d^3+2\right)

\mathrm{Factor}\:3d^3+2,\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(\left(\sqrt[3]{3}\right)^2d^2-\sqrt[3]{2}\sqrt[3]{3}d+\left(\sqrt[3]{2}\right)^2\right)

\mathrm{Refine},\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Hope\:This\:Helps!!!

-Austint1414

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Alex is practicing his dives at a pool. He dies from the diving board that is 15 feet above the surface of water. His dive takes
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Answer:

Alex is wrong in calculating his dive distance Actual Dive distance is 30 feet.

Step-by-step explanation:

Given:

Distance from where he dive from = 15 feet

Distance he reached after dive = 15 feet

Also Given:

Alex says that the two distances are opposites,

Also Total distance of dive according to Alex = 15 +-15 =0 feet

Solution:

We know that the diving board is 15 feet above the water level, but here it is important to know that even he is above the surface of the water he is diving in the water and then he reached 15 deep in the water.

From above scenario we can say that the two distances are in the same direction.

Total Distance of dive will be = 15 +15 =30 feet.

Hence We can say that Alex is wrong in calculating his dive distance Actual Dive distance is 30 feet.

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3 years ago
Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
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Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

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3 years ago
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Answer:

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4 0
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