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Step2247 [10]
3 years ago
5

A drawing of the flag of puerto rico is shown below. Assume that the triangle in the flag is isosceles as shown

Mathematics
1 answer:
torisob [31]3 years ago
4 0
63 

You take 180 subtract 54 and divide it by 2 that gives you the answer.

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Noelle has set a goal to run more than 50 mi this month. So far, she has run 10 mi. Her average running speed is 5 mph.
Vladimir79 [104]

Answer:

To meet her goal, Noelle must run more than 8\ hours the rest of the month

Step-by-step explanation:

Let

h-------> the number of hours Noelle runs for the rest of the month

we know that

The inequality that represent the situation is

10+5h>50

Solve for h

Subtract  10 both sides

5h> 50-10

5h> 40

Divide by 5 both sides

h> 8\ hours

3 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
Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
Reika [66]
N; n+1; n+2 - 3 consecutive numbers

n(n + 1) = (n + 2)² - 19    |use a(b + c) = ab + ac and (a + b)² = a² + 2ab + b²

n² + n = n² + 4n + 4 - 19    |subtract n² from both sides

n = 4n - 15    |subtract 4n from both sides

-3n = -15    |divide both sides by (-3)

n = 5

n + 1 = 5 + 1 = 6

n + 2 = 5 + 2 = 7

Answer: 5; 6; 7.
6 0
3 years ago
Who ever answers gets points
MAXImum [283]

Hello! How are you today?

6 0
3 years ago
List the like terms of the expression 3n+5p+2+n
IgorC [24]

Answer:only the ones that have letterers *m* are like terms

Step-by-step explanation: for example

5j+9u+3+6n+N

5j 9u 6n n are all like terms

4 0
4 years ago
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