Ask question

Ask question

All categories

3 years ago

8

Solve for x . Help me solve them

1 answer:

Tresset [83]3 years ago

8 0

$\\(1).\\\\(1.7898)^{2x} = (1.7898)^6\\\\\implies \ln(1.7898)^{2x} = \ln(1.7898)^6\\\\\implies 2x \ln(1.7898) = 6 \ln(1.7898)\\\\\implies 2x =6\\\\\implies x = \dfrac 62 =3\\\\(3).\\\\2^{5x -6} = 4^{x+9}\\\\\implies 2^{5x -6} = (2^2)^{x+9}\\\\\implies 2^{5x-6} = 2^{2x+18}\\\\\implies \ln(2^{5x-6}) = \ln(2^{2x+18})\\\\\implies (5x -6) \ln 2 = (2x +18) \ln 2\\\\\implies 5x -6 = 2x +18\\\\\implies 5x -2x = 18 +6\\\\\implies 3x = 24\\\\\implies x = \dfrac{24}3 =8$

You might be interested in

Simplify \frac{4(3x^{2}y^{4})^{3}}{(2x^{3}y^{5})^{4}} <br> Show your work.

irina1246 [14]

$\frac{4(3x^{2}y^{4})^{3}}{(2x^{3}y^{5})^{4}} = \frac{4* ( 3^{3} x^{6} y^{12}) }{2^4 x^{12} y^{20}} = \frac{108}{16x^{6} y^{8} } = \frac{27}{4 x^{6} y^{8} }$

4 0

3 years ago

Need help with this!!!!

antoniya [11.8K]

Answer:

Cuatro es la respuesta también conocida como d

6 0

3 years ago

Read 2 more answers

A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L

Scorpion4ik [409]

Answer:

1. $\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

2. $y(40) = 110.873 \ kg$

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should consider y(t) to be the amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that y satisfies can be computed as follows:

$\dfrac{dy}{dt}=rate_{in} - rate_{out}$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{y(t)}{100+2t} \times3$

$\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}$

$\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}$

How much salt is in the tank after 40 minutes?

So,

suppose : $e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}$

Then ,

$( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}$

$( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}$

$[y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}$

Taking the integral on both sides; we have:

$[y(t + 50)^{1.5}] = 1.6 (t + 50)^{2.5} + C$

$y = 1.6 (t+50)+C(t+50)^{-1.5}$

$y(0) = 0 = 1.6(0+50) + C ( 0 + 50)^{-1.5}$

$0 = 1.6(50) + C ( 50)^{-1.5}$

$C= -1.6(50)^{2.5}$

$y(40) = 1.6 (40 + 50)^1 - 1.6 (50)^{2.5}(50+40)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 (90)^{-1.5}$

$y(40) = 144 - 1.6 \times 17677.66953 \times 0.001171213948$

$y(40) = 144 - 33.12693299$

$y(40) = 110.873 \ kg$

6 0

4 years ago

I need to know how to do this please help me out fam

melamori03 [73]

The picture is blurry.

6 0

3 years ago

What is the area of the parallelogram?<br><br><br><br> 80 cm 2<br> 40 cm 2<br> 29 cm<br> 58 cm 2

inessss [21]

Answer:

80cm²

Step-by-step explanation:

Area of parallelogram =base×height

B=16cm

H=5

Area=16×5

80cm²

Hope this helps you :)

8 0

3 years ago

Read 2 more answers