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Natali [406]
3 years ago
15

What is 1.036 that add up to 4

Mathematics
2 answers:
olganol [36]3 years ago
7 0
Answer 2.964

Explanation
Mamont248 [21]3 years ago
6 0

Answer:

2.964

Step-by-step explanation:

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7,14,21,28,35, __
8_murik_8 [283]

Answer:

the patters is that they are multiple of 7.. so next number is 42.

which means option D <u>The missing number is even</u> is correct..

6 0
3 years ago
The population of the Maldives was 390 000 at the start of 2009. The population
My name is Ann [436]

Step-by-step explanation:

start with 390 000x 5.6 x 6

6 0
3 years ago
17/39 + 11/12 −( 5/12 + 4 /39 )
Nana76 [90]

Answer:

.83333333 or 5/6

Step-by-step explanation:

Make common denominator

204/468 + 429/468 = 633/468 - (195/468 + 48/468)

Simplify: 633/468 - 243/468

390/468 can be further broken down to 5/6 if you divide by a common factor which is 78.

8 0
3 years ago
Read 2 more answers
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
3 ten thousand + 14 tens + 16 ones
zzz [600]
30,1416 I'm pretty sure this is the answer
8 0
4 years ago
Read 2 more answers
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