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1 year ago

what is a more accurate method for approximating the volume of the spherical slab other than using just cylindrical slabs.

1 answer:

6 0

A more accurate method for approximating the volume of the spherical slab other than using just cylindrical slabs is the; use the cross sections of the actual sphere, and to not use the cylindrical slabs.

<h3>What is a Cylinder?</h3>

A Cylinder is a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve.

Sphere

This is a regular three-dimensional object in which every cross-section is a circle; the figure described by the revolution of a circle about it's diameter.

Therefore, the use of cross sections of the actual sphere is a more accurate method for approximating the volume of the spherical slab.

Learn more about cylindrical slab:

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Answers fast please. 1.) y=x−63x+2y=8 Use the substitution method. A.) (4, −2) B.) (14, 8) C.) (0, −6) D.) (3, −3) 2.) What is t

GrogVix [38]

<u>QUESTION 1</u>

The given system of equation is

$y=x-6...eqn1$

and

$3x+2y=8...eqn2$

Let us substitute equation (1) into equation (2) to get,

$3x+2(x-6)=8$

We expand the bracket to get,

$3x+2x-12=8$

We simplify to get.

$5x-12=8$

We group like terms to get

$5x=8+12$

$\Rightarrow 5x=20$

$\Rightarrow x=4$

We now substitute $x=4$ in to equation (1) to obtain,

$y=4-6=-2$

The correct answer is option A.

<u>QUESTION 2</u>

The given system of equations is

$y-x=9...eqn1$

and

$10+2x=-2y...eqn2$

We make y the subject in equation (2) to get,

$y=-x-5..eqn3$

We put equation (3) into equation (1) to obtain,

$-x-5-x=9$

We group like terms to get,

$-x-x=9+5$

This implies that,

$-2x=14$

We divide through by $-2$ to get,

$x=-7$

Hence the x-coordinate is $-7$

<u>QUESTION 3</u>

The given system is

$y=3x-1..eqn1$

and

$x-y=-9...eqn2$

We make $x$ the subject in equation (2) to get,

$x=y-9...eqn3$

We put equation (3) into equation (1) to obtain,

$y=3(y-9)-1$

We expand the bracket to get,

$y=3y-27-1$

Group like terms to get,

$y-3y=-27-1$

We simplify to get;

$-2y=-28$

This implies that,

$y=14$

Therefore the y-coordinate is 14.

4 0

3 years ago

Read 2 more answers

HELP QUICK !’ ***What is the area of a circle with a diameter of 126 in?

Vladimir79 [104]

Answer:

49,850.64

Step-by-step explanation:

To find the area of a circle, you must use the formula pi r ^2 (pi r squared).

In this case, we are using 3.14 for pi, so we can add that into the formula.

3.1r^2

We also know the radius, so we can add that in too.

3.14 x 126 ^2

Now solve.

126^2 = 15,876

15,876 x 3.14 = 49,850.64

49,850.64 is the answer.

Hope this helped!

8 0

3 years ago

I need help asap I'll give brainliest

weqwewe [10]

What's the question?

5 0

3 years ago

Help meeeeeeeeeeeeee pleaseeeeeeeeeeeeeeeeeeeeeeeeee

Sati [7]

Answer:

Step-by-step explanation:

8 0

1 year ago

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

3 years ago