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3 years ago

A spherical basketball has a diameter of 18 cm.

Mathematics

2 answers:

lesya [120]3 years ago

3 0

Use the formula 4/3 x pi x r^3 with R being 9
You get 3053.6

forsale [732]3 years ago

3 0

V=(4πr^3)/3 and we are told that d=18cm, so r=9cm thus

V=(4π9^3)/3

V=972π cm^3

V≈3053.63 cm^3 (to nearest hundredth of a cubic centimeter)

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A triangle and a trapezoid are graphed on the coordinate plane shown below.

Gre4nikov [31]

Answer:

10 sq units

Step-by-step explanation:

Area of triangle = 1/2(4)(2) or 4 square units

Area of trapezoid = 1/2(2)(2+4) or 6 square units

Combined area = 10 sq units

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3 years ago

Please help, i really need help

AveGali [126]

Answer :

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Step-by-step explanation:

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3 years ago

What is the area of a square if the base is 4 m in the height is 4 m

iVinArrow [24]

Answer: the area is 16m

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3 years ago

How can you convert the Heun’s Method into the Implicit Heun’s Method? Show an example

g100num [7]

Answer:

Heun's method is also known by its other name called Modified Euler methods. This method is used in computational or mathematical science.

Step-by-step explanation:

Euler method is the method that is also pronounced in two similar stages such as Runge- Kutta methods. This method has been named after Dr. Heun.

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4 0

3 years ago

Solve x^3-7x^2+7x+15

ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

+7x

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

+7x

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

+7x

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

+7(−3)

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

+7(−5)

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

+7x

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0

3 years ago