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3 years ago

What is 26.99 round to the nearest tenth

1 answer:

5 0

Salutations!

To round to the nearest tenth, you need know whether the number next to tenth place is greater than 5 or lesser than 5. If its greater than 5, you need to round up. If its lesser than 5, you need to round down. Now, in the number 26.99 9 is in the tenth place. The number next to 9 is greater than 5, so just add on to the ones place, making the tenth and hundredth place 0. Zero has no value so whether you add the zero or not, it does not matter.

26.99 round to the nearest tenth ≈ 27

Hope I helped (:

Have a great day!

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Solve for y. y y —— = —— P-9 P+9

Veronika [31]

9514 1404 393

Answer:

y = 0

Step-by-step explanation:

Subtracting the right side gives ...

$\dfrac{y}{P-9}-\dfrac{y}{P+9}=0\\\\y\left(\dfrac{(P+9)-(P-9)}{P^2-81}\right)=0\\\\y\left(\dfrac{18}{P^2-81}\right)=0\\\\\boxed{y=0}$

The factor multiplying y cannot be zero, so y must be zero.

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3 years ago

The round off value of 38.23 to the nearest whole number is?

boyakko [2]

Answer:

Step-by-step explanation:

38.23 is closest to 28, because neither the .2 or .03 is high enough to round the last whole number up

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2 years ago

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The jar only contains nickels and dimes altogether there are 10 coins total of the value is 90 cents how many nickels and how ma

lilavasa [31]

Write a system of equations and solve.

x + y = 10
5x + 10y = 90

Multiply and combine equations

-5x - 5y = -50
5x + 10y = 90
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5y = 40
y = 8
x = 2

There are 2 nickels and 8 dimes in the jar. Hope this helps! :)

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3 years ago

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, (P) is the price of the product, (r) is the rate of annual depreciation and (n) 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

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2 years ago

katelyn bought 5.68 pounds of peaches. the peaches were on sale for 1.49 a pound. how much did lately spend on peaches

Leya [2.2K]

So Katelyn bought 5.68 pounds of peaches. And one pound of peaches cost $1.49. So that would mean that we would have to multiply the pounds of peaches she bought, with how much each pound costs. 5.68 × 1.49 = ?

Once you multiply these two decimals, you would get 8.4632. So she spent $8.4632 on 5.68 pounds of peaches. You would multiply the two decimals to find how much she spent in all.

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3 years ago

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