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3 years ago

Find two odd integers whose sum is -88.

Mathematics

2 answers:

AleksAgata [21]3 years ago

6 0

The numbers are -45, and -43
Hope this helps! :D

Sphinxa [80]3 years ago

4 0

X= odd integer, x+2= second odd integer. If their totals equal -88, then:

x+x+2=-88
2x+2=-88
2x=-88-2
2x=-90
x= -90/2= -45, plug into x+2: -45+2= -43
so the numbers would be -45 and -43

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12p+ 15 - 13 - 12p<br><br> And show me how you got the answer

dezoksy [38]

Answer:

Step-by-step explanation:

According to me, you would just have to combine like terms and simplify them as much as possible.

12p + 15 - 13 - 12p

12p - 12p + 15 - 13

0 + 15 - 13

0 + 2

Hope this helps :D

5 0

3 years ago

Three years ago, the annual tuition at a university was $2,000. The following year, the tuition was $2,200, and last year, the t

blondinia [14]

It grows by 10% each year. multiply the number by 0.1, then add. so 0.1×2420 is 242. add 242+2420=2662.

short answer:
this years tuition is $2662.
next years tuition is $2928.

7 0

4 years ago

Read 2 more answers

The width of a rectangle is four units less that its length. Write an expression for the perimeter of the rectangle

Hunter-Best [27]

(2w-8)+2L because there are 2 sides for the width 2*4=8 so 2w -8 plus 2L

5 0

2 years ago

You know that 4 notebooks and 3 pens cost $96, while 2 pens and 2 notebooks cost

bija089 [108]

Answer:

Let p be the number of pens and let n be the number of notebooks.

4 notebooks + 3 pens cost $96 so 4n + 3p = 96

2 notebooks + 2 pens cost $54 so 2n + 2p = 54

We find n and p by solving the system of linear equations:

4n + 3p = 96

2n + 2p = 54

4n + 3p = 96

4n + 4p = 108 (we multiply this by 2 to cancel out 4n in both equations)

We then subtract the two equations:

(4n + 3p) - (4n + 4p) = 96 - 108

-p = -12

p = 12

So, a pen costs 12 dollars. We can use p to find n. We substitute 12 for p in one of our earlier equations:

2n + 2p = 54

2n + 2(12) = 54

2n + 24 = 54

2n = 30

n = 15

So, a notebook costs 15 dollars and a pen costs 12. Now, we need to find the cost of 8 notebooks and 7 pens:

8n + 7p = ?

8(15) + 7(12) = ?

120 + 84 = 204.

Thus, 8 notebooks and 7 pens cost 204 dollars (why is it so expensive lol).

4 0

3 years ago

Read 2 more answers

Given the recursive formula for an arithmetic sequence find the common difference and the 52nd term

Mice21 [21]

Answer:

Common difference(d) $a_{52}$

(21) -10 -548

(22) -7 -323

(23) 10 547

(24) -100 -5118

Step-by-step explanation:

Let the common difference be denoted by 'd'.

Also the nth difference of an arithmetic sequence is given by: $a_{n}=a_{1}+(n-1)\times d$

(21)

We are given a recursive formula as:

$a_{n}=a_{n-1}-10$

The first term is given by:

$a_{1}=-38$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-10$

The nth term of an arithmetic sequence is given by:

$a_{n}=a_{1}+(n-1)\times d$

Here $a_{1}=-38$ and $d=-10$ .

Hence, $a_{52}=-38+(52-1)\times (-10)$

Hence, $a_{52}=-548$

(22)

$a_{n}=a_{n-1}-7$

$a_{1}=34$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-7$

Here $a_{1}=34$ and $d=-7$ .

Hence, $a_{52}=34+(52-1)\times (-7)$

Hence, $a_{52}=-323$

(23)

$a_{n}=a_{n-1}+10$

$a_{1}=37$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=10$

Here $a_{1}=37$ and $d=10$ .

Hence, $a_{52}=37+(52-1)\times (10)$

Hence, $a_{52}=547$

(24)

$a_{n}=a_{n-1}-100$

$a_{1}=-18$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-100$

Here $a_{1}=-18$ and $d=-100$ .

Hence, $a_{52}=-18+(52-1)\times (-100)$

Hence, $a_{52}=-5118$

5 0

4 years ago