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kherson [118]
2 years ago
6

HELPPP!!! ASAPPP!! PLEASE!! WILL GIVE EXTRA POINTS!!

Mathematics
1 answer:
Zina [86]2 years ago
5 0

Answer:

both of those lines intersect each other and are equal lines

Step-by-step explanation:

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Round each number to the nearest ten. What is the best estimate for the answer to 343.42 – 39.87?
Nikolay [14]
So I would round it to 340 and 40 because if the number is lower than 5 than take it DOWN to the nearest ten if it's five or above take UP to the nearest ten

Hope it helped you
7 0
3 years ago
If the cost price of an article is 25% of its selling price , then what is the profit in percentage?
Radda [10]

Answer:

I think 75%.................

4 0
3 years ago
The school day begins at 7:55 a.m. and ends at 2:40 p.m. how long are you in school?
Mademuasel [1]

Think:

7:55 a.m. is 5 minutes (5/60 hrs) before 8 a.m.;

from 8 to noon it's 4 hours, and from noon to 2:40 p.m. is 2 2/3 hours.

Thus, you're in school 5/60 hrs + 4 hrs + 2 2/3 hrs, or

6 2/3 hrs + 5/60 hrs, or 6 40/60 hrs + 5/60 hrs, or

6 45/60 hrs, or 6 3/4 hrs.

Of course there are other ways in which you could do this problem:

4 hrs 5 min plus 2 hrs 40 min comes out to 6 hrs 45 min, or 6 3/4 hrs.

6 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
What is a correct congruence statement for the triangles shown?<br> Enter your answer in the box.
Nesterboy [21]

Answer:

ASA

Step-by-step explanation:

there are two congruent angles with an included side in between

6 0
1 year ago
Read 2 more answers
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