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miss Akunina [59]
3 years ago
8

The area of a gymnasium floor is 264 square yards. the floor is 11 yards wide. How long is the floor?

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

24 yards long

Step-by-step explanation:

if A=l*w then it's 264=l*11

if you divide 264 (area) by 11 (width)  then the answer is 24 which is the length

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8 0
3 years ago
The dimensions of a rectangle are such that its length is 3 in more than its width if the length were doubled and if the width w
emmasim [6.3K]

Length (L): w + 3            ⇒2(w + 3)

width (w): w                    ⇒ w - 1

Area (A) = L x w

         A = (w + 3)(w)

         A = w² + 3w

*******************************************

       A + 176 = 2(w + 3)(w - 1)

(w² + 3w) + 176 = 2(w + 3)(w - 1)

w² + 3w + 176 = 2w² + 4w - 6

        3w + 176 = w² + 4w - 6

                 176 = w² + w - 6          

                   0 = w² + w - 182

                   0 = (w - 13) (w + 14)

0 = w - 13            0 = w + 14

w = 13                  w = -14

Since width cannot be negative, disregard -14

w = 13

Length (L): w + 3   = (13) + 3   = 16

Answer: width = 13 in, length = 16 in

7 0
3 years ago
7. A letter from the word PERSONALITY and a card from a standard deck are chosen at random. How many outcomes are possible?
yKpoI14uk [10]

Answer:

11

Step-by-step explanation:

There are more than eleven cards in a deck but there can only be one outcome because there are only 11 letters in the word PERSONALITY.

8 0
4 years ago
This is part one of the problem posting part two after this
tatiyna

Answer:

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Step-by-step explanation:

If my answer is incorrect, pls correct me!

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7 0
3 years ago
Use polar coordinates to find the volume of the given solid. bounded by the paraboloid z = 5 + 2x2 + 2y2 and the plane z = 11 in
podryga [215]
Using cylindrical coordinates,

\displaystyle\int_{x=0}^{x=\sqrt3}\int_{y=0}^{y=\sqrt{3-x^2}}\int_{z=5+2x^2+2y^2}^{z=11}\mathrm dz\,\mathrm dy\,\mathrm dx
\displaystyle=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=\sqrt3}\int_{z=5+2r^2}^{z=11}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle\frac\pi2\int_{r=0}^{r=\sqrt3}r(11-(5+2r^2))\,\mathrm dr
=\displaystyle\pi\int_{r=0}^{r=\sqrt3}(3r-r^3)\,\mathrm dr
=\dfrac{9\pi}4
3 0
3 years ago
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