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3 years ago

I will give brainiest to whoever answers correctly !!

Mathematics

2 answers:

Zepler [3.9K]3 years ago

8 0

The mark up in the price of the fish dinner is around 50.31%.

evablogger [386]3 years ago

4 0

Answer:

markup=62 %

Step-by-step explanation:

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Which of the following could be the equation of the graph shown below? Check all that apply.

Mashutka [201]

Answer:

y = -2x + 5

Х-y= 9

2x + y = 4

y = 5

lmk if incorrect, please give brainliest

8 0

3 years ago

The two triangles shown below are simiar. Which statement is true of the transformation from ABC to A' B' C'

AfilCa [17]

Answer: Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

Step-by-step explanation:

Since the triangles are similar, the ratio of their sides are equal.

And we can count the number of blocks over which AC and A'C' is drawn and take them to be their length,

Therefore,

AC = 16

A'C'= 8

Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

Measuring the tans of the angles by taking the ratio of opposite by adjacent, we get,

tanA = $\frac{10}{4}$

tanA'= $\frac{5}{2}$

which means tanA= tanA'

The angles do not change.

Thus when transforming from ABC to A'B'C', the lengths are scaled by a factor of 0.5 .

8 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

3 years ago

Which of the following uses the reciprocal property to rewrite {x}{5} = {72}{4}

____ [38]

Anw:

Step-by-step explanation:

72*4=5x

=288

5x=288

x=288/5

=57.6

x=57.6

57.6/72=0.8

4/5=0.8

3 0

2 years ago

Help me with this homework!!!!!!!!!!

Nikolay [14]

Help you with what homework???

6 0

3 years ago