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Bumek [7]
3 years ago
8

On one side of a football stadium are 30 suites on the Circle Level. Fourteen of the suites have 24 seats each, and the rest hav

e 18 seats. Tyler says that 14/30 is the simplest form of the fraction to describe the suites with 24 seats. Is he correct? Use models to support your conclusion and explain why Tyler is correct or incorrect.
Mathematics
1 answer:
ehidna [41]3 years ago
8 0
I’m gonna get to your next week off and I’ll get it to ya next ya know what I can get you guys to join me in my life and I’m gonna go play solo solo for me and I can get it done and I’ll give ya my name
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MAXImum [283]
I believe that x=20 from the math i did hopefully its helpful.
4 0
4 years ago
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Geometry math question no Guessing and Please show work
tatyana61 [14]

The point where the two diagonals meet is also the midpoint of both diagonals.

This means that AE = EC . Since we have the value for both segments, we have

AE = EC \implies 2x = x+2 \implies x = 2

Since AC = AE + EC, we have

AE + EC = 2x+x+2 = 3x+2 = 3\cdot 2 + 2 = 6+2 = 8

4 0
3 years ago
How to solve these equations?
Scorpion4ik [409]
1) / sin x / = y ≥ 0 ;
   We solve the equation y^{2} +y - 2 = 0 ;

a = 1 ; b = 1 ; c = - 2 ;
b^{2} - 4ac = 9 ;
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y2 = ( - 1 - 3 ) / 2 = - 2 ≥ 0 ( false ) ;
Then, / sin x / = 1 ;
sinx = + 1 or sin x = - 1 ;
x ∈ { 90 }  U { 270 } ;
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8 0
3 years ago
\int (x+1)\sqrt(2x-1)dx
Nezavi [6.7K]

Answer:

\int (x+ 1) \sqrt{2x-1} dx =  \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C

Step-by-step explanation:

\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v =  \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}

\int (x+1)\sqrt(2x-1)dx\\\\   = uv - \int v du                              

= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  [ \ u = x + 1 => du = dx  \ ]    

= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\=  \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\

6 0
3 years ago
A bridge is covered with triangular panels to keep the kids from looking down and getting scared. If there are seven panels on e
ICE Princess25 [194]

Answer:

Total area to cover the bridge = 17.52 m²≅

Step-by-step explanation:

The triangular panels are equilateral triangles

Area of equilateral triangles = A= (√3/4)a² where a= 1.7 meters

Since there are 14 panes Area = 14 x  (√3/4)a² = 14 x  (√3/4)1.7² = 17.52 m²≅

4 0
3 years ago
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