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2 years ago

find the volume and surface area of each solid. you may use a calculator and use 3.14 as an approximation for pi

Mathematics

1 answer:

mixas84 [53]2 years ago

3 0

Answer:

Area = 452.16 cm²

Volume = 904.32 cm³

Step-by-step explanation:

The solid is a sphere, therefore,

Volume = ⁴/3πr³

Where,

π = 3.14

r = 6 cm

Volume = ⁴/3*3.14*6³

Volume = 904.319997 ≈ 904.32 cm³ (nearest hundredth)

Area = 4πr²

Area = 4*3.14*6²

Area = 452.16 cm²

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soldier1979 [14.2K]

Answer:

y = 3x

Step-by-step explanation:

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2 years ago

Can Fractions and ratios cannot have zero in the denominator?

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3 years ago

explain why multiplying the numerator and denominator of a fraction by the same number results in an equivalent fraction

Tomtit [17]

Think: You're treating the numerator and the denom. in precisely the same way. In doing so you are NOT changing the value of the fraction, only the appearance.

Example: start with 2/3. Mult num. and den. both by 7: 14/21.

2/3 and 14/21 result in precisely the same decimal fraction, showing that the latter set of fractions is equivalent to the former set.

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3 years ago

Consider the function f(x) = 3x and the function g, which is shown below.

VLD [36.1K]

Answer:

<em>B. The graph of g is the graph of f shifted 2 units down</em>

Step-by-step explanation:

<u>Graph of Functions</u>

We have two functions:

f(x)=3^x

g(x)=3^x-2

Since g(x)=f(x)-2 it will be represented as an identical graph as that for f(x), but vertically displaced 2 units down. Let's check it by plugging some points

f(0)=3^0=1

g(0)=3^0-2=-1

f(1)=3^1=3

g(1)=3^1-2=1

f(3)=3^3=27

g(3)=3^3-2=25

We can notice the values of g(x) are always 2 units below f(x), thus the correct answer is

B. The graph of g is the graph of f shifted 2 units down

7 0

3 years ago

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

3 years ago