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Romashka-Z-Leto [24]
3 years ago
10

(This is literally 7th grade math dude should be easy for most of you)

Mathematics
2 answers:
Tasya [4]3 years ago
8 0

no b. 40.05

a. to figure out the number of goals per game you have to divide the total amount of goals by the number of games:

last season: 41÷ 15 = 2.73

this season: 24 ÷ 9 = 2.67 =

as you can see the answer to the last season equation is higher than the this season meaning the player scored more goals last season.

b. we already know that the amount of games the player won each game (all of them supposedly being exactly even) is 2.67. so what we have to do is multiply it by 15 the supposed number of games the player is playing this season to get the total number of goals they will get for the entire season:

2.67 x 15 = 40.05 = the prediction of the total goals of the player

ANSWER
Korolek [52]3 years ago
7 0

Answer:

a. no b. 40.05

Step-by-step explanation:

a. to figure out the number of goals per game you have to divide the total amount of goals by the number of games:

last season: 41 ÷ 15 = 2.73

this season: 24 ÷ 9 = 2.67

as you can see the answer to the last season equation is higher than the this season meaning the player scored more goals last season.

b. we already know that the amount of games the player won each game (all of them supposedly being exactly even) is 2.67. so what we have to do is multiply it by 15 the supposed number of games the player is playing this season to get the total number of goals they will get for the entire season:

2.67 x 15 = 40.05 = the prediction of the total goals of the player

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Step-by-step explanation:

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The sphere of the radius = x^2 + y^2 +z^2 = 4^2

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z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

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Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

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dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

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x = rcosθ = x; x varies from 0 to 2π

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dA = rdrdθ

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The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
3 years ago
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