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tatyana61 [14]
3 years ago
10

486 divided by 19 i need the remainder with answer

Mathematics
2 answers:
zavuch27 [327]3 years ago
8 0
It’s 25.578 but rounded to the tenths place would be 26.0
Goshia [24]3 years ago
5 0

Answer:

25 remainder 11

Step-by-step explanation:

used a calc

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What is the volume of a cylinder that is 40 inches deep and 16 inches in diameter
Rudik [331]
In a cylinder
height,h=40inches
diameter,d=16inches
radius,r=16/2
=8inches

so, by formula,

volume,v= πr^2h
=22/7*8*8*40
=8045.71 cubic inches
5 0
4 years ago
The graph represents the heights of two climbers on a climbing wall over a 12-minute time period.
pychu [463]

Both climbers rested on the wall at a constant height for 2 minutes.

<h3>How to determine the true statement?</h3>

The illustrative graph that represents the climbers' movement is added as an attachment

From the attached graph, we have the following observation

  • The line of Abby's movement is horizontal from x = 4 to x = 6
  • The line of Brynn's movement is horizontal from x = 5 to x = 7

The difference between the endpoints of both horizontal line is

Difference = 6- 4 = 2

Difference = 7 - 5 = 2

This means that both Brynn and Abby stopped for 2 minutes

Hence, the true statement about is that (d) Both climbers rested on the wall at a constant height for 2 minutes.

Read more about graphs at:

brainly.com/question/4025726

#SPJ1

8 0
2 years ago
Suppose that g(x) = f(x) + 2. Which statement best compares the graph of g(x) with the graph of f(x)?
Jlenok [28]

Answer:

<u>A</u>

Step-by-step explanation:

Since it mentions g(x) is f(x) + 2, it means the y-intercept of g(x) is 2 more than that of f(x).

This basically means that :

<u><em>The graph of g(x) is shifted 2 units up.</em></u>

6 0
2 years ago
Select from the drop down menus to correctly complete each statement
Sav [38]
Sentence 2: left I don't know if that quailifys as a answer...
5 0
3 years ago
Read 2 more answers
Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where
mel-nik [20]

Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :

[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint

1 + 8/<em>n</em> (<em>i</em> - 1)

Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height f(x_i) :

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)

Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}

3 0
3 years ago
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