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3 years ago

Please help! Will award brainliest. Simplify the expression -5+i/2i. Please show your work.

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

5 0

(-5+i)/2i

multiply by i/i

(-5+i) *i = -5i +i^2 = -5i -1

2i*i = 2i^2 = 2(-1) = -2

(-1-5i)/-2

1/2 +5/2 *i

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A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation $A=45e^{-0.05t}$ .

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate $A=15$ and solve for t as:

$15=45e^{-0.05t}$

$\frac{15}{45}=\frac{45e^{-0.05t}}{45}$

$\frac{1}{3}=e^{-0.05t}$

Now we will switch sides:

$e^{-0.05t}=\frac{1}{3}$

Let us take natural log on both sides of equation.

$\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})$

$-0.05t\cdot (1)=\text{ln}(\frac{1}{3})$

$-0.05t=\text{ln}(\frac{1}{3})$

$t=\frac{\text{ln}(\frac{1}{3})}{-0.05}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}$

$t=-\text{ln}(\frac{1}{3})\cdot 20$

$t=-(\text{ln}(1)-\text{ln}(3))\cdot 20$

$t=-(0-\text{ln}(3))\cdot 20$

$t=20\text{ln}(3)$

Therefore, it will take $20\text{ln}(3)$ years for area of the glacier to decrease to 15 square kilo-meters.

6 0

2 years ago

Radioactive Waste The rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt,

Dmitry_Shevchenko [17]

Answer:

$\displaystyle{\int^{\infty}_0 {50e^{-0.04t}}\, dt}=1250$

Step-by-step explanation:

Given:

$T =\displaystyle{\int^{\infty}_0 {Pe^{-kt}} \, dt}$

where,

T = total amount of waste

P = 50, the initial rate

k = 0.04

t = time

$T =\displaystyle{\int^{\infty}_0 {50e^{-0.04t}} \, dt}$

now we need to solve this integral!

$T =\displaystyle{50\int^{\infty}_0 {e^{-0.04t}} \, dt}$

$T = \left|50\left(\dfrac{e^{-0.04t}}{-0.04}\right)\right|^{\infty}_0$

$T = \left|-1250e^{-0.04t}\right|^{\infty}_0$

$T = (-1250e^{-0.04(\infty)})-(-1250e^{-0.04(0)})$

when any number has a power of negative infinity it is 0. because: $a^-{\infty} = \dfrac{1}{a^{\infty}} = \dfrac{1}{\infty} = 0$ , like something being divided by a very large number!