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3 years ago

What is the radius of a sphere that has a surface area of 196π sq ft?

Mathematics

1 answer:

grandymaker [24]3 years ago

3 0

4πr^2=196π r^2=49 r=7

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Find the surface area of the shape below.

almond37 [142]

Answer:

54m^2

Step-by-step explanation:

Each square has a equal sides, which is 3m

in order to find the surface area, it's

Formula : Base x height = Area

The total surface area would be all the area's added together

3x3=9cm^2

so each square is 9m^2

there are 6 squares so

9+9+9+9+9+9=54m^2

or you can do 9x6=54m^2

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Evaluate the following: (½)(10) −(−12) −25

mr Goodwill [35]

Answer:

-8

Step-by-step explanation:

1/2 times 10 is 5. and a negative times a negative is a positive, so you have a positive 12, which becomes 5+12-25, which is -8

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2 years ago

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73 foot showed.......................

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2 years ago

Solve for x 1.6 equals 12.8

xenn [34]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe

wariber [46]

Answer:

(a)

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

(b)

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

(c)

$(A - B) - C = \{a\}$

$A - (B - C) = \{a,b,c\}$

They are not equal

Step-by-step explanation:

Given

$A= \{a,b,c\}$

$B =\{b,c,d\}$

$C = \{b,c,e\}$

Solving (a):

$A\ u\ (B\ n\ C)$

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ (A\ u\ C)$

$A\ u\ (B\ n\ C)$

B n C means common elements between B and C;

So:

$B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}$

$B\ n\ C = \{b,c\}$

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}$

u means union (without repetition)

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

Using the illustrations of u and n, we have:

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C$

Solve the bracket

$(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C$

Substitute the value of set C

$(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}$

Apply intersection rule

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C)$

In above:

$A\ u\ B = \{a,b,c,d\}$

Solving A u C, we have:

$A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}$

Apply union rule

$A\ u\ C = \{b,c\}$

So:

$(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

The equal sets

We have:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

So, the equal sets are:

$(A\ u\ B)\ n\ C$ and $(A\ u\ B)\ n\ (A\ u\ C)$

They both equal to $\{b,c\}$

So:

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

Solving (b):

$A\ n\ (B\ u\ C)$

$(A\ n\ B)\ u\ C$

$(A\ n\ B)\ u\ (A\ n\ C)$

So, we have:

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})$

Solve the bracket

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})$

Apply intersection rule

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}$

Solve the bracket

$(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}$

Apply union rule

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})$

Solve each bracket

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}$

Apply union rule

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

The equal set

We have:

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

So, the equal sets are:

$A\ n\ (B\ u\ C)$ and $(A\ n\ B)\ u\ (A\ n\ C)$

They both equal to $\{b,c\}$

So:

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

Solving (c):

$(A - B) - C$

$A - (B - C)$

This illustrates difference.

$A - B$ returns the elements in A and not B

Using that illustration, we have:

$(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}$

Solve the bracket

$(A - B) - C = \{a\} - \{b,c,e\}$

$(A - B) - C = \{a\}$

Similarly:

$A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})$

$A - (B - C) = \{a,b,c\} - \{d\}$

$A - (B - C) = \{a,b,c\}$

They are not equal

4 0

3 years ago