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2 years ago

A breakfast stall at a market stays

Mathematics

1 answer:

zheka24 [161]2 years ago

6 0

Answer:

If it opens at 5:30, then it closes at 10:00 AM

If it opens at 6, then it closes at 10:30 AM

If it opens at 6:30, then it closes at 11:00 AM

If it opens at 7, then it closes at 11:30 AM

If it opens at 7:30, then it closes at 12:00 PM

After 12:00 PM, no one is going to have breakfast. So these are the only possibilities of the timing it will open. Either at 5:30, 6, 6:30, 7, or 7:30.

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Solve for m -mk-90>85

loris [4]

First add 90 to both sides) -mk>85+90simplify 85+90 to 175) -mk>175divide both sides by K) -m>175/kmultiply both sides by -1) m<-175/k
YOUR ANSWER IS
M<-175/K
hope l helped today l was stuck on the same question doing part 2 of alg exam , and l came through your question l hope it's not too late :)

5 0

3 years ago

Witch equation represents a linear function

olya-2409 [2.1K]

Y=3x2+1
Hope this helped :)

5 0

2 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago

1. Find the domain and range of the function<br> f(x)=√1-4x².

slamgirl [31]

$f(x) = \sqrt{1 - 4{ x}^{2} }$

$1 - 4x {}^{2} \geqslant 0$

$- \infty \: \: \: \: \: \: \: \: - 0.5 \: \: \: \: \: \: \: \: \: 0.5 \: \: \: \: \: \: \: \: \infty \\ - \: \: \: \: \: \: \: \: \: \: \: \: \: \:0 \: \: \: \: \: + \: \: \: \:0 \: \: \: \: \: -$

$domain \\ [ \: -0.5 \: , \: 0.5 \: ]$

$g(x) = 1 - 4x {}^{2} \\ g'(x) = - 8x \\ g'(x) = 0 \\ x = 0 \\ g(0) = 1 - 4(0) = 1 \\ maximum \: = 1$

$range \\ [ \: 0 \: , \: 1 \: ]$

5 0

2 years ago

Read 2 more answers

24 students and Jamal's class are wearing tennis shoes there are 30 students in this class your mom says that 70% of this class

mel-nik [20]

Im not sure if i am understanding the question correctly...

If you are asking if Jama's mom is correct, then no. she is wrong. this is because

if you take the proportion, (is/of)=(%/100) you get (x/30)=(70/100). You then solve for x which in this case would be 21 students are wearing tennis shoes.

Hope this helps.

3 0

3 years ago