# What is 8.546 rounded to the nearest tens?

10

Step-by-step explanation:

In this problem we need to round off 8.546 to the nearest tens. 8 is in ones place, 5 is in tenths place, 4 is in hundredths place and 6 is in thousandth place.

8 falls between 5 and 10 on the number line. It is less than 5 and more than 10. When it is rounded off to nearest tens it can be given by 10.

To round 8.546 to the nearest tens, the answer is 10.

To round 8.546 to the nearest tens, you need to look at the digit in the ones place, which is 6.

Since 6 is greater than or equal to 5, you round up the tens digit, which is 4.

Therefore, the final answer is 10.

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## Related Questions

Find each percent increase. Round to the nearest percent

1. 15 to 21 = 21 - 15 / 15 = 0.4*100% = 40%

2. 12 teachers to 48 teachers = 48 - 12/48 = 0.75 * 100% = 75%

3. 80 pencil to 120 pencil = 120 - 80 / 80 = 0.5*100% = 50%

4. 40 cans to 70 cans = 70 - 40 / 70 = 0.43 * 100% = 43%

This is the two that go together, but are separate questions

Option 1)

6a + 8s = 102

14a + 4s = 150

each adult ticket costs 9 dollars

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Explanation:

6 adult tickets and 8 student tickets bring in $102, so that means 6a+8s = 102 14 adult tickets and 4 students tickets bring in$150, so 14a+4s = 150

The system of equations is

If we multiply both sides of the second equation by -2, we get this updated system

Add the equations straight down

6a+(-28a) = -22a

8s+(-8s) = 0s = 0 ... the 's' terms go away

102+(-300) = -198

So we end up with the equation -22a = -198 and that solves to a = 9 after dividing both sides by -22.

Each adult ticket costs $9 If you want the value of s, then 6a+8s = 102 6(9)+8s = 102 54 + 8s = 102 8s = 102-54 8s = 48 s = 48/8 s = 6 Meaning each student ticket costs$6

Or you could use the other equation

14a+4s = 150

14(9)+4s = 150

126+4s = 150

4s = 150-126

4s = 24

s = 24/4

s = 6

We get the same value of s

3 cm3 cm
3 cm
2 cm
3 cm
6 cm
Find the surface area of the above solid.
A. 81 cm2
B. 78 cm2
C. 84 cm2
D. 72 cm2

Step-by-step explanation:

Suppose a professor splits their class into two groups: students whose last names begin with A-K and students whose last names begin with L-Z. If p1 and p2 represent the proportion of students who have an iPhone by last name, would you be surprised if p1 did not exactly equal p2? If we conclude that the first initial of a student's last name is NOT related to whether the person owns an iPhone, what assumption are we making about the relationship between these two variables?

a) Even if the distribution of iPhones by last name is completely uniform in the population generally, there is no reason to believe that the proportions in the sample represented by the class will be identical.
I would not be surprised to see p1 ≠ p2.

b) Saying the variables are not related is the same as saying the variables are independent.

A.3(f) The line graphed on the grid represents the first of two equations in a system of linear equations.20
-16
12
-8
-20 -16 -12
-
-4
48
12
16 2024
-8
-12
16
If the graph of the second equation in the system passes through (-12, 20) and (4,12), which statement is true?

I don’t any idea for this

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients.y'' − y' − 2y = 5e−t

y = C₂ e²ᵗ + e⁻ᵗ [C₁ - (5t/3)]

Step-by-step explanation:

y" - y' - 2y = 5e⁻ᵗ

The total solution of this differential equation will be a sum of its complimentary and particular solutions.

y = y꜀ + yₚ

The complimentary or homogenous solution is a solution of

y" - y' - 2y = 0

y꜀ = Ay₁ + By₂

Solving this part in the attached image, we obtain

y₁ = e⁻ᵗ and y₂ = e²ᵗ

The particular solution can be obtained using the method of variation of parameters and the method of undetermined coefficients.

The method of variation of parameters is used and explained in the page 1, 2 and page 3 of the attached images.

The method of undetermined coefficients is then used to check if the solution obtained is correct.

With y = C₂ e²ᵗ + e⁻ᵗ [C₁ - (5t/3)],

y' and y'' were obtained and put in the equation, and the answer obtained was 5e⁻ᵗ.

So, Proved!

y = C₂ e²ᵗ + e⁻ᵗ [C₁ - (5t/3)]

Hope this helps!!!