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3 years ago

Alex and DJ go to the movies at 5:15. The movie lasts 2 hours and 25 minutes. What time did they get out of the move?

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

Answer:

7:40

Step-by-step explanation:

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What is the area of this figure 3,7,7

IRINA_888 [86]

I think the answer is 147

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3 years ago

Is 5.692 terming or rrepeating

german

Terminating, because it ends. If it was repeating it would go in infinitely, like Pi.

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3 years ago

The school baseball team has 3 pitchers. How many different 3-player pitching schedules can the coach choose from?

expeople1 [14]

Answer:

6 schedules

Step-by-step explanation:

First, decide if this situation describes a combination or a permutation by looking at the significance of order. Consider a specific example where the letters A – C represent the 3 different pitchers. The pitching order A, B, C is different from the pitching order B, C, A because the players will be pitching at different times. Therefore, this situation describes a permutation. Use the formula for finding the number of permutations when the number of pitchers is 3, so n = 3, and all 3 are taken at a time, so r = 3 as well.

7 0

3 years ago

Find it nearest tenths

NNADVOKAT [17]

Answer:

Step-by-step explanation:

finding PR ( public relations :D , not that PR , the PR on this triangle thingy )

use law of cosines, again

c=sq rt [ a^2+b^2﹣2*a*b*cos(C) ]

where

c=PR

a= 9

b = 22

C = the angle of Q or 180 = 97+24+Q so Q = 180-97-24 = 59°

they tried to trick us.. by not giving us the angle C for the above cosine formula , but we figured out their diabolical plan :D yay super hero math

then

PR = sq rt [ 9^2 + 22^2 - 2*9*22*cos(59) ]

PR = sq rt [ 81 + 484 -203.9550777 ]

( notice that now it is a subtraction sign instead of a plus sign, this time? )

PR = sq rt [361.0449 ]

PR = 19.0011

to the nearest 10th

19.0 :)

6 0

2 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

Alex and DJ go to the movies at 5:15. The movie lasts 2 hours and 25 minutes. What time did they get out of the move?​

Alex and DJ go to the movies at 5:15. The movie lasts 2 hours and 25 minutes. What time did they get out of the move?