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3 years ago

Sales tax is 10.25%. What is the tax on a 300$ bill?

Mathematics

1 answer:

Liula [17]3 years ago

7 0

$30.75
is taxes sale total is 330.75

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Write a function rule for your bacterium in the table where h is the number of hours, and b(h) is the bacterium at h hours.

kolbaska11 [484]

Answer:

B(h)=x (subscript)-1*4+1 basically x from the previous generation times 4 plus one.

Step-by-step explanation:

5 0

3 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

Someone please help asap, I will mark as brainliest. Thank you!!

defon

Answer:

the answer is c go ahead and click it

Step-by-step explanation:

8 0

2 years ago

I need help on this question

Irina18 [472]

To find the surface area of the figure , we need to add the surface area of all the faces

So the surface area of figure = Area of two lateral rectangular faces + Area of two triangular faces + area of base

hence

Surface Area of figure = 2(10*15) + 2( 1/2 * 8 * 12) + (12*15)

= 300 + 96 + 180

= 576 square feet

So correct option is the last one

7 0

3 years ago

How do I solve this?

Crazy boy [7]

Answer:

71°

142°

Step-by-step explanation:

$\because QS \: is \: the \: bisector \: of \: \angle \: PQR\\\therefore m\angle RQS =m\angle PQS\\\because m\angle RQS = 71\degree...(given) \\ \huge \red{ \boxed{\therefore m\angle PQS = 71\degree}} \\ \because \: m\angle \: PQR = m\angle PQS + m\angle RQS \\ \because \: m\angle \: PQR = 71\degree + 71\degree \\ \huge \purple{ \boxed{\therefore \: m\angle \: PQR = 142\degree}}$

7 0

3 years ago