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3 years ago

PLEASE HELP Find the surface area of the triangular prism.

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

139

Step-by-step explanation:

First, find all the faces that you must find the areas of. In this case, you need to find the areas of two triangles and three rectangles.

In this triangular prism, the base is one triangle. If the area of the base is 7.75, we can assume that that holds true for both bases, and multiply 7.75 by 2 in order to find the area of both triangles.

Now find the area of each of the three rectangles by multiplying their individual heights and bases by each other. You should get 38, 47.5, and 38 as the areas of the rectangles.

Now add all the individual area's together. (They're bolded for clarity).

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A department store sells logo shirts at an original price of $20. Every month that a shirt doesn’t sell, the store reduces the s

Tasya [4]

Given:
Original price = 20
reduces selling price by 25% every month it's not sold.

First markdown month:
20 * (100%-25%) = 20 * 75% = 15

Second markdown month
15 * 75% = 11.25

Macy, employee gets a 50% discount off the current price.
11.25 * 50% = 5.625
11.25 - 5.625 = 5.625 or 5.63

The pre-tax price of the shirt for Macy will be $5.63

3 0

3 years ago

Read 2 more answers

HELP ME PLEASE <br> PLEASE PLEASE PLEASE PLEASE PLEASE

Nimfa-mama [501]

Answer:

19y+5

Step-by-step explanation:

Combine like terms look for all the numbers that have y attached and combine them

7y +12y=19y

Look for terms where nothing is attached it's just the number combine them

6-1=5

Put all together again and you get:

19y+5

8 0

3 years ago

An adult soccer league requires a ratio of at least 2 women per 7 men on the roster. If 14 men are on the roster, how many women

Snezhnost [94]

Answer:

Atleast 4 women

Step-by-step explanation:

Ratio of

Women to men = 2 : 7

Number of women needed to maintain the ratio if there are 14 men on the roster :

The minimum number of women required :

(2 : 7) * number of men in roster

(2 / 7) * 14

2 * 2 = 4 women

Atleast 4 women are required to main the ratio

7 0

3 years ago

Use double integrals to calculate the area inside the ellipse whose semiaxes have length a and b

Stells [14]

The general equation of an ellipse centered at the origin with its semiaxes coinciding with the coordinate axes is given by

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Substituting $x=a\cos\theta$ and $y=b\sin\theta$ into the above equation gives the Pythagorean identity, so we can use polar coordinates quite nicely to our advantage.

If $\mathcal E$ is the ellipse with the equation above, the area is given by the double integral

$\displaystyle\iint_{\mathcal E}\mathrm dA=\iint_{(x/a)^2+(y/b)^2\le1}\mathrm dx\,\mathrm dy$

Let $x(r,\theta)=ar\cos\theta$ and $y(r,\theta)=br\sin\theta$ , so that the Jacobian matrix is

$\mathbf J=\begin{bmatrix}x_r&x_\theta\\y_r&y_\theta\end{bmatrix}=\begin{bmatrix}a\cos t&-ar\sin t\\b\sin t&br\cos t\end{bmatrix}$

and the magnitude of its determinant is $|\det\mathbf J|=|abr|=abr$

since in polar coordinates we use the convention that $r\ge0$ , and $a,b>0$ because they are lengths.

Now, the area is given by

$\displaystyle\iint_{\mathcal E}\mathrm dA=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}abr\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi a b\int_{r=0}^{r=1}r\,\mathrm dr$
$=\pi a b$

4 0

4 years ago

A chain 27 feet long whose weight is 95 pounds is hanging over the edge of a tall building and does not touch the ground. How mu

bazaltina [42]

Answer:

Work done $= 2,565$ lb-ft

Step-by-step explanation:

The density of the chain in lb/ft is equal to

$\frac{95}{27}$ lb/ft

Total Work done is equal to the summation of work done up to the top of building.

We will use integration for this evaluation.

The mathematical relation is as follows

$\int\limits^{40}_0 {X} \, pdx$

where "p" is the density

Substituting the given values, we get

$\frac{95}{27} \int\limits^{27}_0 {X} \, dx\\\frac{95}{27} X^2\int\limits^{27}_0\\ \frac{95}{27} {27^2 - 0^2}\\\frac{95}{27} * 27 * 27\\95* 27\\= 2,565$

Work done $= 2,565$ lb-ft

6 0

3 years ago