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brilliants [131]
3 years ago
13

875,932,461,160 what digit is in the hundred-millions place of this number.

Mathematics
1 answer:
coldgirl [10]3 years ago
8 0

The digit in the hundred millions place is 9.

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Answer this please thanks
Nesterboy [21]

Answer:

175 per day

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What is the answer to the question
-BARSIC- [3]

Answer:

From top to bottom:

3

7

6

12

Step-by-step explanation:

You just simply add the number in the top row and the number in the column together.

2 + 1 = 3

4 + 3 = 7

2 + 4 = 6

6 + 6 = 12

5 0
2 years ago
Indicate the method you would use
FinnZ [79.3K]

Answer:

correct choice is 1st option

Step-by-step explanation:

Two given triangles have two pairs of congruent sides: one pair of length 7 units and second pair of length 8 units. The third side is common, i.e the lengths of third sides are equal too.

Use SSS theorem that states that if three sides of one triangle are congruent to three sides of another triangle, then these two triangles are congruent.

Thus, given triangles are congruent by SSS theorem.



7 0
3 years ago
Let f be defined by the function f(x) = 1/(x^2+9)
riadik2000 [5.3K]

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

converges by comparison to the convergent <em>p</em>-series,

\displaystyle\sum_{n=3}^\infty\frac1{n^2}

(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

converges absolutely, since

\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0
3 years ago
Please explain how this digraph works.
dusya [7]

Angles XQL and MQR are congruent because they are vertical angles. So

209 - 13 <em>b</em> = 146 - 4 <em>b</em>

Solve for <em>b</em> :

209 - 13 <em>b</em> = 146 - 4 <em>b</em>

209 - 146 = 13 <em>b</em> - 4<em> b</em>

63 = 9 <em>b</em>

<em>b</em> = 63/9 = 7

Then the measure of angle XQL is

(209 - 13*7)º = 118º

3 0
3 years ago
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