A bike path is 6 3/4 miles long. What whole number is closest to 6 3/4?

HELP ITS HARD PLZZZ need help to 1 throw 6

Nata [24]

Ok so here's a trick king Henry doesn't usually drink chocolate milk or king Henry dabs under dogs calling mom and here's what it will look like the pic and then u have to convert them so 8 liters converting into blank dL so u put a x over the letter L and u don't see that. u see the big U well that means gram liter and meter so u put a x over top of that and now u want to go to the d so now u go just to the d NOT da and ur gonna scoop when ur done scooping them u put a dot where u ended on top of that letter now u get ur first number or what one u started with so I did 8L so I get 8 and since there is no decimal u put it at the end and now I'll scoop and since ur getting smaller u move the decimal to the left (cause left is less) and u get .8 and that's how u do that if u need help with the other ones just let me know and I'll help u

3 0

3 years ago

Read 2 more answers

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago

Puntuación de una Línea determinada?

taurus [48]

Answer:

score of a certain Line?

Step-by-step explanation:

If you wanted a translation, that's your answer!

5 0

3 years ago

(a) Fiona has $450 in her savings account. She deposits $40 each month. Liam has $975 in his checking account. He writes a check

kenny6666 [7]

Answer:

y=450+40x y=975-65x

Step-by-step explanation:

at five months they will have the same amount of money in their accounts

7 0

3 years ago

HELPPPPPPPPP i will answer one of your question if you answer

Alika [10]

Its red because 25+7(1)=25+7=32, 25+7(2)=25+14=39, and so on

3 0

3 years ago