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3 years ago

The surface area of a triangular pyramid is 375.2 m2. What would change if the surface area were measured in square centimeters?

1 answer:

8 0

I believe that D. The surface area of the pyramid would stay the same, but the number representing the surface area would increase is correct because when you just move it to another units, it still the same but you'll need to move the decimal two places to the left and the number 37520 cm² is greater than the number of 375.2 m²

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on monday 3.11 inches of rain fell and on tuesday 0.81 inch of rain fell. on wensday,twice as much rain fell as tuesday. How muc

pshichka [43]

Answer:

5.54 inches of rain

Step-by-step explanation:

Monday - 3.11

Tuesday - 0.81

Wednesday - 0.81 x 2= 1.62

3.11 + 0.81 + 1.62=5.54 inches

4 0

2 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

-8.46 greater or less than 5.31

Gekata [30.6K]

Hello.

-8.46 is less than 5.31, because it is negative.

4 0

3 years ago

Read 2 more answers

The sales tax rate is

ruslelena [56]

The price will be $25.30

7 0

3 years ago

Math answer plz need help with this plz answer

Neporo4naja [7]

Answer:

176.625 sq.ft, 4.8 sq.ft

Step-by-step explanation:

Area of circle=πr^2 or πd^2/4
11. Given,
d=15 ft
Now,
Area=πd^2/4
3.14*15^2/4
176.625 ft^2
Therefore, the area approximation is 176.6 sq.ft
12.
Given,
d=3.5 ft
Now,
Area= πd^2/8
3.14*3.5^2/8
4.8 sq.ft
I got the answer by dividing the area of circle by 2 as semicircle is half of circle.

6 0

2 years ago