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3 years ago

Multiplication made for 840 here's the numbers 11 3 4 5 2 3 7 6 14 what do I use to get 840 using only those numbers Help quick!

1 answer:

7 0

you can use the numbers 3, 4, 5, 2, 6, 7, and 14. You can multiply 3 times 280 to get 840, 210 times 4 to get 840, 5 times 168 to get 840, 2 times 420 to get 840, 140 times 6 to get 840, 7 times 120 to get 840, and 14 times 60 to get 840.

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Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71. $\overline {6}$ °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

$\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)$

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

$\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min$

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

$\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min$

8 0

3 years ago

A plant that is 6 inches tall grows 1 inch each week. Another plant is 4 inches tall and grows 2 inches each week

Darya [45]

2 weeks, because the first week the second plant will grow to 6 inches and the first plant will grow to 7, the next week the first plant will be 8 inches and so will the second plant.

4 0

2 years ago

Explain why all linear angle pairs must be supplementary, but all supplementary angles do not have to be linear pairs?

dimulka [17.4K]

Two angles are supplementary if their measures sum 180°.

Linear angle pairs are adjacent angles (they are together) and they form a straight angle, which is 180°. Therefore, they are supplementary.

But two angles may be separated and yet add 180°, then they are supplementary but are not linear angle pairs.

4 0

4 years ago

Could someone explain how they get the answer for this?

jarptica [38.1K]

Answer:

40°

Step-by-step explanation:

The reference angle is the positive acute angle created by the terminal arm and the x-axis.

The highlighted red in the picture below shows what we're looking for.

The arm rotated 220° (but 'backwards' so the value given is negative).

|-220°| - 2(90°) <= Subtract two right angles for two quadrants

= 220° - 2(90°)

= 220° - 180°

= 40°

Therefore, the reference angle is 40°.

If you got 50°, you probably calculated the angle with the terminal arm and the y-axis. Remember to always use the nearest side of the x-axis!

4 0

4 years ago