All categories

3 years ago

HELP ME, PLZ. 100 POINTS !!!

Mathematics

1 answer:

marta [7]3 years ago

3 0

Answer:

x=2√5/5

x=-2√5/5

Step-by-step explanation:

13√(2√5/5)^2-(2√5/5)^4+9√(2√5/5)^2+(2√5/5)^4=16

13√(4×5/25)-(16×25/625)+9√(4×5/25)+(16×25/625)

13√(4/5)-(16/25)+9√(4/5)+(16/25)

13√(4/25)+9√(36/25)

13×2/5+9×6/5

26/5+54/5

16=16 (proven)

You might be interested in

A $1508 award is shared equally by 8 people. What is each persons share of the award

andreev551 [17]

Answer:

$188.50

Step-by-step explanation:

i got this answer by dividing $1,508 by 8 and you would get 188.5 which would turn into $188.50 and that is how much each person would get.

I hope this helps.

6 0

3 years ago

Read 2 more answers

Evaluate:<br> (a)<br> (8 x 8) - 24

Alexus [3.1K]

Answer:

Step-by-step explanation:

i solved for you

hope it helped! ;)

6 0

3 years ago

Read 2 more answers

Are segments PQ and NM parallel?

katen-ka-za [31]

Answer:

no, i dont think so because parralel lines are this

Step-by-step explanation:

hope it helps

8 0

2 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

High beam headlights must not be used within

faust18 [17]

What’s The Question Again ?

7 0

3 years ago