Ask question

Ask question

All categories

3 years ago

15

Distribute and simplify these radicals. I need help!

2 answers:

Iteru [2.4K]3 years ago

8 0

Answer:

4 V (square root) 3

Step-by-step explanation:

notka56 [123]3 years ago

3 0

Answer:

The last one (4 sqare root 3)

Step-by-step explanation:

The square root of 16 is 4 so then it would be 4 sqroot 3

You might be interested in

PLEASE HELP ME QUICKKK, FIRST CORRECT PERSON GETS BRAINLIEST

Vinil7 [7]

A b and d hope it helps

7 0

2 years ago

Read 2 more answers

In the figure, the horizontal lines are parallel and AB = BC = CD. If LM = 3 Find JM? A. 12 B. 6 C. 3 D. 9

ziro4ka [17]

A.12 because i had same question

<span />

5 0

3 years ago

TEA [102]

Answer:

B. -0.5

Step-by-step explanation:

I calculated it logically

4 0

3 years ago

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

3 years ago

by 14 1 4 each week for a number of weeks, w . He writes the expression 20(14) 20 ( 1 4 ) w to find the amount of radioactive su

erastovalidia [21]

Answer:

1/2 ^w : decay factor after w weeks

hope this helps

Step-by-step explanation:

8 0

3 years ago