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Bingel [31]
3 years ago
13

How many minutes between 12:53 and 1:47

Mathematics
2 answers:
Yuki888 [10]3 years ago
8 0
54 minutes. .....................

melomori [17]3 years ago
4 0
60 min minus 6 =54 min
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What is 615 rounded to the nearest hundred?
Svetlanka [38]
615 ≈ 600 (nearest hundred)
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4 years ago
Twice the smallest of three consecutive odd integers is nine more than the largest. find the integers.
polet [3.4K]
 let   x        x +2       x+4     three consecutive odd <span>integers


2x =9+ x+4   </span>Twice the smallest<span> is nine more than the largest 
 2x - x= 13

x= 13  the first 
the second integer 15   the third  17 </span>
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3 years ago
How do u solve y=3x 5x+2y=44
Digiron [165]

Answer:  which will give you 15x+3y=44 and you cannot do anything else because you don't have any like terms any more so that's your answer.

Step-by-step explanation: the first thing you need to do first is to combine like terms. and if i did a mistake sorry but i hope this helps.

4 0
3 years ago
Point C is twice as close to point B as it is to point A. If point A is at -30 and point B is at 30, what are two possible value
arsen [322]

different between A and B

=30-(-30)

=60

position of C from A or B

=(60÷3)×2

=40

possible value of c

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8 0
4 years ago
Read 2 more answers
Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1
PolarNik [594]

Answer:

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

Step-by-step explanation:

We are given that function

f(x)=(1+3x)^{-1/3}

We have to find the  first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....

f(0)=(1+3x)^{\frac{-1}{3}}=1

f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}

f'(0)=-1

f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}

f''(0)=4

f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}

f'''(0)=-28

Substitute the values we get

(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...

(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

5 0
3 years ago
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