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3 years ago

Dividing a decimal by a decimal 1.1÷1.21

Mathematics

2 answers:

shepuryov [24]3 years ago

7 0

The answer is 1.1
new to this ok

QveST [7]3 years ago

5 0

Answer-1.1
How and Why- 1.1 has 1 decimal moved over so on the inside you move the decimal point one place over the right. 1.1 goes in 12 only one time. Subtract then you have just 11. Which goes 1 time. So your answer would be 1.1

Hope this helps

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The sum of two consecutive integers is 7. find the numbers

faust18 [17]

X + (x + 1) = 7

2x + 1 (-1) = 7 (-1)

2x = 6

2x/2 = 6/2

x = 3

x +1 = 3 + 1

x + 1 = 4

The two integers are 3, 4

hope this helps

4 0

3 years ago

Read 2 more answers

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Charra [1.4K]

Answer:

- 2 - 3i

Step-by-step explanation:

given a complex number a + bi

then the conjugate is a - bi

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3 years ago

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lora16 [44]

Answer

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Step-by-step explanation:

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6 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago