Ask question

Ask question

All categories

2 years ago

9

Pls pls pls pls pls help

1 answer:

Arturiano [62]2 years ago

7 0

Answer:

$y=\frac{1}{2}$

Step-by-step explanation:

$4(y+1)= \frac{3}{1-y}$

$4y+4= \frac{3}{1-y}$

$(4y+4)(1-y)= 3$

$4y(1-y)+4(1-y)=3$

$4y-4y^{2} +4-4y=3$

$-4y^{2}=3-4$

$4y^{2}=1$

$y^{2} = \frac{1}{4}$

$y=\sqrt{\frac{1}{4} }$

$y=\frac{1}{2}$

You might be interested in

Solve the following proportion 12/25=x/88

Fittoniya [83]

<span>12/25=x/88
12(88)=25(x)
1056=25x divide both sides by 25
x=42.24</span>

8 0

2 years ago

Spiral Review Elena draws a scale drawing for a garden in her backyard. The scale drawing is 3 inches by 6 inches. The garden wi

faust18 [17]

1/4 bc if you take the fraction 3/12 and simply it’s 1/4

3 0

3 years ago

How to solve burger barn has 46 2/3 pounds of ground beef. how many 1/3 pound burgers can be made using all the ground beef?

8_murik_8 [283]

(46 2/3) / (1/3) =
(140/3) / (1/3) =
140/3 * 3/1 =
420/3 =
140 <=== 140 can be made

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

2 years ago

The yearly attendance at a local movie theater is 68,000 and grows continuously at a rate of 3.5% each year what is the approxim

asambeis [7]

$3.5 \times 8 - 68000$
the answer is 67972

7 0

3 years ago

Read 2 more answers