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3 years ago

I need help with step 1 and step 2

Mathematics

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:

x=.1460

Step-by-step explanation:

ln5^(2x+1)=ln8

2x+1ln5=ln8

2x+1=(ln8/ln5)

x=((ln8/ln5)-1)/2

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The monthly rent of Marilyn's house went from $500 to $400, if p is the percent decrease in the rent, which proportion can be us

faust18 [17]

For this case we can use the following rule of three:
$ 500 ---------> 100%
$ 400 ---------> x
Clearing x we have:
x = (400/500) * (100) = 80%
Therefore, the percentage of decrease in the rent is:
p = 100-x
p = 100-80
p = 80%
Answer:
a proportion can be used to calculate p is:
p = 100 - ((400/500) * (100))

3 0

2 years ago

How to write3.repeating 7 as a mixed number

ki77a [65]

Hey there! I'm happy to help!

Let's call our fraction x.

x=3.777......

We want to get rid of the repeating stuff. If we multiply our fraction by 10, we get 10x=37.7777.

We see that if we subtract x from 10x, the repeating numbers will be cancelled out. Let's do this.

10x=37.7777

x=3.77777

9x=34

We divide both sides by 9.

x=34/9=3 7/9.

Have a wonderful day! :D

6 0

3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$