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3 years ago

Which expression is equivalent to 3 sqrt (x^5y)

Mathematics

1 answer:

Marat540 [252]3 years ago

7 0

Answer:

$3x^{2}\sqrt{xy}$

Step-by-step explanation:

we have

$3\sqrt{x^{5}y}$

we know that

$\sqrt{x^{5}} =\sqrt{x*x^{4}}=x^{2}\sqrt{x}$

substitute

$3\sqrt{x^{5}y}=3x^{2}\sqrt{xy}$

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Given f(x) = x3 – 2x2 – x + 2, <br><br> the roots of f(x)

anyanavicka [17]

f(x) = x³ – 2x² – x + 2

0 = x3 – 2x2 – x + 2

(x-2)(x-1)(x+1)=0

x1=-1

x2=1

x3=2

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3 years ago

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Nataly_w [17]

they are equal or congruent because if you look at it they both are right triangles

4 0

4 years ago

Any answers I need help

natulia [17]

Answer:

(A) 21

Step-by-step explanation:

5x-15 needs to equal 90 since it is a right angle and 21 makes that true

$(5*21)=105\\105-15=90$

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2 years ago

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If x-y=9 and xy=2,find the value of (x square + y square)

kati45 [8]

Answer:

x² + y² = 85

Step-by-step explanation:

Using the expansion

(x - y)² = x² + y² - 2xy , then

x² + y² - 2xy = (x - y)² ( add 2xy to both sides )

x² + y² = (x - y)² + 2xy ← substitute given values

= 9² + 2(2)

= 81 + 4

= 85

3 0

3 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago