Bob works as a carpenter and needs to buy some tools for

How to solve 3n-4=11

Vlad1618 [11]

Here u go i hope this correct

5 0

3 years ago

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Need help ASAP give brainest

satela [25.4K]

4-3 3-1 2-2 1-4 ( the first side I numbered 1-4 and the other side I numbered 1-4)

7 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

Please help with this no fakes

I am Lyosha [343]

9514 1404 393

Answer:

h ≤ -7/2

Step-by-step explanation:

This is what is known as a "2-step" problem.

The first step is to "isolate the variable term" by adding the opposite of the term on the left that does not have the variable in it. That is, we want to add -4 to both sides of the inequality.

4 -4 +2h ≤ -3 -4 . . . . add -4 to both sides

2h ≤ -7 . . . . . . . . . . variable term isolated

Now, we want to get rid of the coefficient of the variable (make it be 1), so we use the multiplicative inverse. That is, we multiply both sides of the inequality by 1/2, the reciprocal of the coefficient.

(1/2)(2h) ≤ (1/2)(-7) . . . . . multiply both sides by 1/2

h ≤ -7/2 . . . . the solution

__

For an inequality, multiplying or dividing by a negative number causes the direction of the comparison to be reversed. (Here, our multiplier was 1/2, not negative, so no such reversal was required.)

<u>Example</u>:

1 < 2

-1 > -2 . . . . . both sides multiplied (or divided) by -1

_____

<em>Additional comment</em>

As when solving equations, you must do the same thing to both sides of the inequality.

3 0

3 years ago

Which of the following is a rotation? Select all.

MakcuM [25]

Answer:

A B C D

Step-by-step explanation:

Hey you said to select all

A and D

5 0

3 years ago

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