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masha68 [24]
3 years ago
9

Evaluate 8j-k+14when j=0.25 and k =1

Mathematics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

8j-k+14

8(0.25)-1+14=2-1+14=15

Step-by-step explanation:

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100 POINTS!!!!!!! I WILL GIVE BRAINLIEST!
abruzzese [7]

Answer:

  • 8.4 miles

Step-by-step explanation:

Let the speed in the morning be s, then in the evening it is s - 4

<u>The distance is equal both directions:</u>

  • 45 min *s = 1 hr 10 min *(s - 4)

<u>Time in hours:</u>

  • 45*1/60*s = (1 + 10*1/60)(s - 4)
  • 3/4s = 7/6s - 7/6*4

<u>Multiply all terms by 12 to clear fraction:</u>

  • 9s = 14s - 56
  • 5s = 56
  • s = 56/5
  • s= 11.2 mph

<u>The distance is:</u>

  • 3/4*11.2 = 0.75*11.2 = 8.4 miles
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3 years ago
If x plus one upon X equal to 11 find the value of x square + 1 upon x square
zubka84 [21]
119 <answer (check the solution in above pic)


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3 years ago
Sin(73) = cos (x) pls help lol
viva [34]
X = 17+360n, 343+360n
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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.

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3 years ago
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ArbitrLikvidat [17]

Go to Settings > [Your Name] > iTunes & App Store.

Tap your Apple ID account and select View Apple ID.

Sign in with Face ID, Touch ID, or your Apple ID password.

Tap Subscriptions, then select an active subscription.

Tap Cancel Subscription

4 0
3 years ago
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