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xxTIMURxx [149]
3 years ago
15

Find the sum of 914 and 878

Mathematics
2 answers:
irga5000 [103]3 years ago
6 0
1792
~~~~~~~~~~~~~~~~~
Cerrena [4.2K]3 years ago
3 0
The word "sum" means addition, so the question is asking you to add 914 and 878.

First you can add the hundreds, so 900 + 800 = 1,700
Then you can add the tens, so 10 + 70 = 80
And finally the units, so 4 + 8 = 12
When you add up all of these, you get your final answer as 1,792.

I hope this helps!
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rob is looking into a new cell phone plan to see if he could save money. plan a which he already has consists of a $45 fee for a
Dimas [21]

Plan A:

$45+$.30t=$150

Plan B:

$60+$.25t=$150

7 0
3 years ago
What is the solution of the system?
kakasveta [241]

Divide the first equation by 2 and add the result to the second equation. This will eliminate x.

... (-4x -2y)/2 + (2x +4y) = (-12)/2 +(-12)

... 3y = -18 . . . . . collect terms

... y = -6 . . . . . . . divide by 3

Substitute this into either equation to find x. Let's use the second equation, where the coefficient of x is positive.

... 2x +4(-6) = -12

... 2x = 12 . . . . . . . . add 24

... x = 6 . . . . . . . . . . divide by 2

The solution is (x, y) = (6, -6).

8 0
3 years ago
How many positive integers between 1000 and 9999 inclusive
bekas [8.4K]

Answer:

Step-by-step explanation:

a.

first number is  1000-1+9=1008

9)1000(1

    9

-------

     10

       9

    -----

       10

         9

       ----

         1

       ----

last number is 9999

9| 9999

  ---------

    1111 |0

    --------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

  ---------

  |4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total  numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

  ---------

  |  142-6

1000-6+7=1001

7 | 9999

  |---------

    1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

        70

       ----

        300

        280

        ------

           20

           ---

1000-20+35=1015

35)9999(285

     70

    ----

     299

     280

     -----

        199

        175

        ----

          24

         ----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

7 0
3 years ago
A grocer mixed pecans worth $0.84 per pound with cashews worth $1.40 per pound. If he made 88 pounds of a mixture worth $1.26 pe
Vsevolod [243]

Answer:

There are 22 pounds of pecans in the mixture

Step-by-step explanation:

Let the number of pounds of pecans be p and the number of cashews be c

We have 88 pounds in total, mathematically, this means;

c + p = 88

Now, the 88 mixtures cost $1.4 per pounds. This means the total amount of the 88 would be 88 * 1.26= $110.88

The pecans are worth 0.84 per pound while the cashews are worth 1.4 per pound. If the total is at a cost of 110.88, this can be translated mathematically as:

0.84p + 1.4c = 110.88

We now have two different equations that we can solve simultaneously.

c + p = 88

0.84p + 1.4c = 110.88

From the first equation, we can say c = 88 - p

We can then substitute the into the second equation:

0.84p + 1.4(88 - p) = 110.88

0.84p + 123.2 - 1.4p = 110.88

-0.56p = 110.88 - 123.2

0.56p = 12.32

p = 12.32/0.56 = 22

Now, we know c = 88 - p = 88 - 22 = 66

4 0
3 years ago
Read 2 more answers
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Makovka662 [10]

Answer:

Step-by-step explanation:

For the points lol

5 0
3 years ago
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