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3 years ago

Subtract the sum of -11ab and -23ab from the sum of -19ab and 7ab

1 answer:

7 0

The sum of -19ab and 7ab = -19ab + 7ab = -12ab
the sum of -11ab and -23ab = -11ab + (-23ab) = - 34ab

So
Subtract the sum of -11ab and -23ab from the sum of -19ab and 7ab:

 (-12ab) - (-34ab) = -12ab + 34ab = 22ab

Answer
22ab

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I solved this problem : 33 + 45 + 78 how could i check my answer?

Vlad [161]

Step-by-step explanation:

So the answer for 33+45+78 is 156.

If you want to check that, you can use inverse operations, so subtract.

156-78=78. (Check 78)

78-45=33 (Check 45 and 33)

Well, now it's checked because we used all the factors of the equation!

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3 years ago

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Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621

REY [17]

Answer:

The Sum Of The Integers From -6 To 58 is 1690.

Step-by-step explanation:

Given,

$a=-6$

$T_n=58$

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here $a_1=-6\ and\ a_2=-5$

$\therefore d=a_2-a_1=-5-(-6)=-5+6=1$

Now we use the formula of A.P.

$T_n=a+(n-1)d$

On substituting the values, we get;

$58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65$

So there are 65 terms in between -6 To 58.

That means we have to find the sum of 65 terms in between -6 To 58.

Now we use the formula of Sum of n_terms.

$S_n=\frac{n}{2}(2a-(n-1)d)$

On substituting the values, we get;

$S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690$

Hence The Sum Of The Integers From -6 To 58 is 1690.

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Step-by-step explanation:

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Svet_ta [14]

Answer:

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Step-by-step explanation:

Given data

Let y represent the total amount of money

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We can model this situation as

y= 450x

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Answer:

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