Jorge went to Target to buy a Lego Kit. The Lego Kit is worth $25.00 plus a 8.5% sales tax. How much is Jorge paying in sales tax?

Recall that

[tex]\dfrac{d}{dx}\left[\cot(x)\right] = -\csc^2(x)[/tex]

so that

[tex]\displaystyle \int \frac{dx}{\sin^2\left(\frac{x-2}3\right)} = \int \csc^2\left(\frac{x-2}3\right) \, dx = \boxed{-3 \cot\left(\frac{x-2}3\right) + C}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{Your \ exercise \ is \to\int\limits \frac{1}{sin^2\left(\dfrac{x-2}{3}\right) }dx } \end{gathered}$}[/tex]

             [tex]\boldsymbol{Replaces \ u=\dfrac{x-2}{3} \longmapsto \dfrac{du}{dx}=\dfrac{1}{3} \longmapsto \ dx=3 \ du }[/tex]

                                               [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=\int\limits \frac{3}{Sin^2(u) }du } \end{gathered}$}[/tex]

                                                    [tex]\boldsymbol{\sf{Simplify }}[/tex]

                                              [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=3\int\limits csc^2(u)du } \end{gathered}$}[/tex]

                                                    [tex]\boldsymbol{\sf{Solving \ now }}[/tex]

                                                    [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\int\limits csc^2(u)du } \end{gathered}$}[/tex]

                                            [tex]\boldsymbol{\sf{This\:is\:a\:standard\:integral. }}[/tex]

                                                     [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-cot(u)} \end{gathered}$}[/tex]

                                       [tex]\boldsymbol{\sf{We\:replace\:the\:integrals\:already\:solved. }}[/tex]

                                                   [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=\int\limits \frac{3}{Sin^2(u) }du } \end{gathered}$}[/tex]

                                                      [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-cot(u)} \end{gathered}$}[/tex]

                                      [tex]\boldsymbol{The\:substitution\:is\:undone\: u=\dfrac{x-2}{3} ;}[/tex]

                                                      [tex]\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{=-3cot\left(\frac{x-2}{3}\right) } \end{gathered}$}[/tex]

       [tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf Answer \longmapsto \bf{\int\limits \dfrac{1}{sin^2\left(\dfrac{x-2}{3}\right) }dx }=\boldsymbol{ -3cot\left(\frac{x-2}{3}\right)+C } \end{gathered}$}}[/tex]

[tex]\large\displaystyle\text{$\begin{gathered}\sf \underline{\bf{\green{I\:hope \ it \ helps \ you..... Regards }}} \end{gathered}$}[/tex]

Answer:

1.28!

Step-by-step explanation:

hope this helps, let me know if you need an explination! :)

Answer:

Step-by-step explanation:

Reason:

Triangles A, B and E all have the angles 68 and 79 with a 3 unit side sandwiched between them. We can use the angle side angle (ASA) congruence theorem to prove those three triangles are identical copies of each other. They are rotated, reflected and shifted versions of one another. These tranformations are known as rigid transformations or isometries.

On the other hand, we can rule out choice C because the side between the angles is not 3 units long (even though the angles do match up). The same goes for choice D.

Answer:1) In Δ FGH,

FH=7 ft, FG=12 ft,

∠F=70 degrees

By "Law of cosines":

So,By applying Law of Sines , we get that

Hence, Option 'D' is correct.

2) In Δ XYZ,

XY=14, ∠Y=22°, and XZ=26

By using "Law of sines":

so, X = 147° , Y = 22° , Z = 11.6°

Step-by-step explanation:

Answer:

360

Step-by-step explanation:

9x4=36 add the zero you didn't use

The expression that returns the probability that a data value in a normal distribution is between a z- score of -0.18 and a z-score of 1.99 is (d) normalcdf(-0.18, 1.99)

To calculate the probability between two z-scores is to calculate the area between the z-scores.

This is done using the following syntax:

normalcdf(lower bound, upper bound)

From the question, the lower and the upper z-scores are -0.18 and 1.99

So, the formula would be normalcdf(-0.18, 1.99)

Hence, the expression is (d)

Read more about normal distribution at:

https://brainly.com/question/25991460

Answer:

1,005cm²

Step-by-step explanation:

First, you need to know the area of a cylinder and also the value of pi.

Area = 2πrh

pi = π = 3.14

r = radius = 80cm

h = height of cylinder = 20

now let's substitute

[tex]2 \times 3.14 \times 80 \times 20 = 1004.9{cm}^{2} [/tex]

but to the nearest square centimetre is.

1005cm²

because the number after the point is more than 4 so we add one to the last number before the point.

The angle of fire in the standard position will be B. 120°; 2 pi over 3 radians.

It should be noted that the image shows the shooting angles arrangements. Also, π = 180°.

In this case, the value of the angle will be:

= 12π/18

= (12 × 180) / 18

= 12 × 10

= 120°

Therefore. the angle of fire in the standard position will be B. 120°; 2 pi over 3 radians.

Learn more about angles on:

https://brainly.com/question/25770607

Applying proportions, it is found that the discounted price of each console at store B is given by:

B. $210.80.

A proportion is a fraction of a total amount.

In this problem, at store B, 15 consoles sold for a total of $3,720, hence the price per console is of:

P = 3720/15 = $248.

A discount of 15% is applied, which means that the new price is 85% of the original price, hence:

P = 0.85 x 248 = $210.8.

Which means that option B is correct.

More can be learned about proportions at https://brainly.com/question/24372153

Answer:

[tex]\tan(X)=\dfrac{77}{36}[/tex]

Step-by-step explanation:

First we need to find the length of side WX.

To do this, use Pythagoras' Theorem:  [tex]a^2+b^2=c^2[/tex]

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

[tex]\implies WX^2+77^2=85^2[/tex]

[tex]\implies WX^2=85^2-77^2[/tex]

[tex]\implies WX=\sqrt{85^2-77^2}[/tex]

[tex]\implies WX=36[/tex]

Now use the Tan trig ratio:

[tex]\tan(x)=\sf \dfrac{O}{A}[/tex]

where:

From inspection of the triangle,

Substituting these values into the tan trig formula:

[tex]\implies \tan(X)=\dfrac{77}{36}[/tex]

Therefore, the tangent of ∠X is [tex]\dfrac{77}{36}[/tex]

Answer:

77 / 36

Step-by-step explanation:

Finding the adjacent side

Tangent of ∠X

Answer:

[tex]1590.6cm {}^{3} [/tex]

Step-by-step explanation:

[tex]volume \: of \: a \: cone \: v = \frac{1}{3}\pi \: r {}^{2} h \\ radius = \frac{diameter}{2} = \frac{15}{2} = 7.5 \\ v = \frac{1}{3} \times 3.42 \times (7.5) {}^{2} \times 27 \\ v = \frac{3.142 \times 7.5 \times 7.5 \times 27}{3} \\ v = \frac{4771.9125}{3} = 1590.6375 \\ v = 1590.6cm {}^{3} (approximatly) \\ please \: rate \: and \: mark \: brainliest[/tex]

Answer:

4,000

Step-by-step explanation:

The value goes down by around 2,000 every year and 4,00 is the closest answer of any of them.

Given that

[tex]\vec A = xz^3 \, \vec\imath - 2x^2 y \, \vec\jmath + 2yz^4 \, \vec k[/tex]

its curl is

[tex]\displaystyle \nabla\times\vec A = \left(\frac{\partial\left(2yz^4\right)}{\partial y} - \frac{\partial\left(-2x^2y\right)}{\partial z}\right) \, \vec\imath - \left(\frac{\partial\left(2yz^4\right)}{\partial x} - \frac{\partial\left(xz^3\right)}{\partial z}\right) \, \vec\jmath \\ ~~~~~~~~~~~~ + \left(\frac{\partial\left(-2x^2y\right)}{\partial x} - \frac{\partial\left(xz^3\right)}{\partial y}\right) \, \vec k[/tex]

[tex]\nabla\times\vec A = 2z^4 \, \vec\imath + 3xz^2 \, \vec\jmath - 4xy \, \vec k[/tex]

so that at the point (1, -1, 1), the curl is

[tex]\nabla\times\vec A \bigg|_{(x,y,z)=(1,-1,1)} = \boxed{2 \, \vec\imath + 3 \, \vec\jmath + 4 \, \vec k}[/tex]

Answer:

It shows a deficit of $261

Step-by-step explanation:

$32,508/12 = $2709 per month

$2709 - ($1860 + $860 + $259) = -$270

I'm assuming you put a typo somewhere, but the closest answer is the deficit of $261

We can represent the pattern by a general sequence:

u(n+1)=8n+2

u(1)=8(10)+2

u(1)=82

u(2)=8(82)+2

u(2)=658

....

Answer:

10:82:658:5266:42130

Answer:

$5.25

Step-by-step explanation:

here,

4pounds off apple cost $7

3 pounds of apple will cost 3×7/4= 21/4=$5.25

Answer:

10.6 oz

Step-by-step explanation:

X × 0.25 = 0.265

x= 2.65/0.25= 10.6 oz

Answer:

1. 5/(2+2)= 5/4= 1.25

2. 5/(2x2)= 5/4= 1.25

Step-by-step explanation:

3x5/12= 15/12= 1.25

Answer:

Step-by-step explanation:

A number decreased by 16 is − 3

hope this helps.

mark me brainliest please.

Answer:

10

Step-by-step explanation:

here it is

the explanation is in the photo

[tex]v (total)= v(r \: p \: ) + v(c)[/tex]

[tex]v(t) = lwh + \pi {r}^{2} h[/tex]

[tex]v(t) =( 18 \times 10 \times 5 )+( 3.14 \times {5}^{2} \times 5) \\ [/tex]

[tex]v(t) = 900 + 392.50[/tex]

[tex]v(t) = 1292.50 \: \: \: {ft}^{3} [/tex]

Depth is also referred as height.

Volume of the rectangular part = l b h

= 18 × 10 × 5

= 900 ft³

Volume of two semi circular parts = 4 / 3 π r³

= 4 / 3 × 5 × 5 × 5 × π

= 523.8 ft³

Total Volume of the figure = 1423.8 ft³

Answer:

x times square then subtract

Answer:

1.91, 1.92, 1.93

Step-by-step explanation:

They are all between 1.9 and 2

Answer:

B Building = 865 feet,   A Building = 665 feet.

Step-by-step explanation:

If you divide 1530 by 2 you get 765, if you add 100 to 765, and subtract 100 from 865, there is a 200 feet difference from each of the buildings.

Answer:

I hope it helps you

Answer:

Step-by-step explanation:

The graph of the linear equation cuts the x-axis at (5,0)

Here's your Question →

____________________________________

Que → The hieght of The statue of Liberty is 93m. This is about 27% of the Hancove building in Chicago. What is approximate Hieght of Hancove Building?

Answer → Let height of Hancove Building be H.

Now according to question, 27% of H = 93m

27 / 100 (H) = 93m

H = 93 x 100 ÷ 27

H = 344.44m

Therefore, the approximate Hieght of Hancove Building is around 344.4m

Answer:

Step-by-step explanation:

Let the missing height is h.

We have 93 m is 27% of h, need to find the value of h

27% of h is 0.27h, the equation we need is below