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Alik [6]
3 years ago
10

What is 2160 minus 1600 long way

Mathematics
1 answer:
lilavasa [31]3 years ago
8 0
Hey there, Lets solve this problem together. 

The First step is to line up the numbers. 

=  2160

- 1600 

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left} 

<span>We calculate </span>0-0<span>the result of which is </span>0<span> 
</span>
<span>We calculate </span>6-0<span> the result of which is </span>6<span>.
</span>
Since we get a negative number in the next column, we must take 1 from the next column and carry it over to this column. Now the number will be changed to 10.

We calculate 10+1-6, and the result is 5. 

<span>We calculate </span>2-1-1<span> the result of which is</span>0<span>. 
</span>
Therefore, 

2160 - 1600 = 0560 
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Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, i
Doss [256]
C. Hope this helps :)
5 0
3 years ago
Sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).
Sever21 [200]
\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta

\sin\alpha=\dfrac{21}{29}\implies \cos^2\alpha=1-\sin^2\alpha=\dfrac{400}{841}

Since \alpha lies in quadrant II, we have \cos\alpha, so

\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}

\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}

\beta lies in quadrant I, so \sin\beta>0 and

\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}

So

\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}
8 0
3 years ago
Whats the answer for 4 and 5? Please I need the answers ASAP.
olga nikolaevna [1]

Answer:

4. 6/15

Step-by-step explanation:

5. 5/3 is the an answer

8 0
2 years ago
What is the least common demoninator for the fractions
vlabodo [156]

Answer:

The least common denominator is 6.

Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
(4y − 3)(2y2 + 3y − 5)
VladimirAG [237]

Answer:

8y^3+6y^2-29y+15

Step-by-step explanation:

the correct expression is

(4y − 3)(2y^2 + 3y − 5)

Given data

We have the expression

(4y − 3)(2y^2 + 3y − 5)

let us open bracket

8y^3+12y^2-20y-6y^2-9y+15

Collect like terms

8y^3+12y^2-6y^2-20y-9y+15

8y^3+6y^2-29y+15

4 0
3 years ago
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