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IrinaK [193]
2 years ago
8

Solve: (a+b+c) (a+b-c)

Mathematics
2 answers:
Katen [24]2 years ago
7 0

Answer:

<h2><u><em>a²+2ab+b²-c²</em></u></h2>

Step-by-step explanation:

Solve:

(a+b+c) (a+b-c)=

(a²+ab-ac+ab+b²-bc+ac+bc-c²)=

a²+ab-ac+ab+b²-bc+ac+bc-c²=

a²+2ab+0ac+b²+0bc-c²=

a²+2ab+b²-c²

zvonat [6]2 years ago
4 0

Answer:

The answer is a² + 2ab + b² - c²

Step-by-step explanation:

(a + b + c) × (a + b - c)

a² + ab - ac + ba + b² - bc + ca + cb - c²

a² + ab + ba - ac + ca + b² - bc + cb - c²

a² + 2ab + b² - c²

Thus, The answer is a² + 2ab + b² - c²

<u>-TheUnknownScientist</u><u> 72</u>

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-4+x= -11 please answer it
Slav-nsk [51]

Answer:

x=-7

Step-by-step explanation:

add 4 to both sides. leaving you with x=-7

8 0
3 years ago
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What does this expression mean when it says 2 less than five times a number
Zinaida [17]

Answer:

2 - 5x

Step-by-step explanation:

Answer is in the problem:

2 less than = 2 minus a number ( 2- )

five times a number = 5x ( we don't know what number five is being multiplied by, so we put a variable)

6 0
3 years ago
a farmer uses 1/3 of his land to plant cassava 3/5 of the remaining land to plant Maize and the rest vegetables if vegetables co
Rzqust [24]

Answer:

37.5 acres.

Step-by-step explanation:

Fraction of his land  where vegetables are grown

= 1 - 1/3 - 2/3 *3/5

(after taking away 1/3 we get 2/3 and the farmer uses 3/5 of this 2/3 to grow Maize.

= 1 - 1/3 - 6/15              LCD of 3 and 15= 15  so we write:

= 1 - 5/15 - 6/15

= 1 - (5+6)/16

=  1 - 11/15

= 4/15.

4 /15 is equivalent to  10 acres

so  total  area (= 1)  =   10 / 4/15     Invert the 4/15 and multiply:

= 10 * 15/4

=  150/4

= 37.5 acres.

3 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
Four time the sum of five and ten is what
frozen [14]
The answer is sixty!
4(5+10)
4(15)
60
3 0
3 years ago
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