All categories

3 years ago

Which digit is in the thousands place in 385,419?

Mathematics

2 answers:

Georgia [21]3 years ago

6 0

The digit in the thousands place is 5

Anton [14]3 years ago

5 0

The digit is 5..........

You might be interested in

What are the answers

kolezko [41]

Answer:

I use app called math

way

Step-by-step explanation:

7 0

2 years ago

A city planner designs a park that is a quadrilateral with vertices at J(-1, 1), K (3, 3), L (5, -1), and

Liono4ka [1.6K]

9514 1404 393

Answer:

83 m

Step-by-step explanation:

The attached diagram shows the lengths of the midlines to be 3.16 and 5.10 units on the coordinate plane. The sum of these is (3.16 +5.10) = 8.26 coordinate plane units. Since each unit on the coordinate plane represents 10 m, the total path length is about ...

8.26(10 m) = 82.6 m ≈ 83 m

_____

The distance formula will tell you the distances are ...

AC = √(3² +1²) = √10

BD = √(5² +1²) = √26

The total path length is about (10 m)(√10 +√26) ≈ 82.61230 m ≈ 83 m.

3 0

3 years ago

Find the center and radius of the circle whose equation is x 2 + y 2 + 10x − 1

Margaret [11]

we conclude that the center of the circle is the point (-5, 0).

<h3>How to find the center of the circle equation?</h3>

The equation of a circle with a center (a, b) and a radius R is given by:

$(x - a)^2 + (y - b)^2 = R^2$

Here we are given the equation:

$x^2 + y^2 + 10x - 1 = 0$

Completing squares, we get:

$x^2 + 10x + y^2 = 1$

Now we can add and subtract 25 to get:

$x^2 + 10x + 25 - 25 + y^2 = 1\\\\(x + 5)^2 -25 + y^2 = 1\\\\(x + 5)^2 + (y - 0)^2 = 26$

Comparing that with the general circle equation, we conclude that the center of the circle is the point (-5, 0).

If you want to learn more about circles:

brainly.com/question/1559324

#SPJ1

3 0

2 years ago

When factored completely, the expression p^4-81 is equivalent to what?

Mice21 [21]

Differnce of 2 pefect squares
a^2-b^2=(a-b)(a+b)
(p^2)^2-9^2=(p^2-9)(p^2+9)

p^2-9=(p-3)(p+3)

factored is
(p-3)(p+3)(p^2+9)

4 0

3 years ago

Read 2 more answers

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$