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3 years ago

Help me please: (√5)^2 x (√3)^2

Mathematics

2 answers:

Bezzdna [24]3 years ago

5 0

Answer:

Step-by-step explanation:

Note that $\sqrt{a}$ × $\sqrt{a}$ = ( $\sqrt{a}$ )² = a

Thus

( $\sqrt{5}$ )² × ( $\sqrt{3}$ )² = 5 × 3 = 15

yawa3891 [41]3 years ago

4 0

Answer:

Step-by-step explanation:

$(\sqrt[]{5} )^{2} = 5$

$(\sqrt{3} )^{2} = 3\\$

$5$ x $3 = 15$

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Rama09 [41]

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3 years ago

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Fantom [35]

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3 years ago

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The perimeter of a triangle is 32 feet. One side of the triangle is 3 times the second side. The third side is 12 feet longer

Ivenika [448]

We have a triangle whose perimeter is 32 feet. Some information regarding the sides of the triangle is given:

GiveN:

One side is 3 times the second side.
Third side is 12 feet longer than the second side.

As we can see that, two sides of the triangle are defined the second sides. It means First and third side can be expressed in the form of second side.

So let the second side be x

Then,

First side = 3x
And, third side = x + 12

We know that, perimeters is the sum of the lengths of all sides of the triangle, So it can be written as:

$3x + x + x + 12 = 32$

Solving it further,

$5x + 12 = 32$

Subtracting 12 from both sides,

$5x + 12 - 12 = 32 - 12$

$5x = 20$

Dividing 5 from both sides,

$\frac{5x}{5} = \frac{20}{5}$

$x = 4$

Then,

First side = 3x = 12 feet
Second side = 4 feet
Third side = x + 12 = 16 feet

And we are done with the answer !!

#CarryOnLearning

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3 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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4 years ago

Can you help please no links or point stealing or I will report

valentina_108 [34]

1.A school assembly has 179 students in it. Nine teachers escort "k" number of

students out of the assembly, until there are 26 students remaining.

2. a i guess

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3 years ago

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