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NNADVOKAT [17]
2 years ago
11

The sum of two number is 50,and their difference is 30.find the numbers​

Mathematics
2 answers:
Mrac [35]2 years ago
6 0

Answer:

40 and 10

Step-by-step explanation:

if you add them you get 50 and if you subtract them you get 30

rodikova [14]2 years ago
3 0

Answer:

40 and 10

Step-by-step explanation:

Let the bigger number = x

Let the smaller number = y

x+y=50

x-y=30

(add the equations)

(x+x)+(y-y)=(50+30)

2x=80

x=40

x+y=50

40+y=50

y=10

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Directions: Match the product and quotients estimates below with the correct expression on the left
prohojiy [21]

Answer:

Answer is in attached image.

Step-by-step explanation:

Given the expressions, for which we have to find the estimates as per the expressions on the left.

The given expressions are:

1) 35 \times 23

2) 132 \div 168

3) 17.3 \times 18.4

4) 999 \div 208

5) 998 \times 211

Here, we need to find the rounded off numbers.

35 can be rounded to 40 and 23 to 20.

Therefore, equivalent to 35 \times 23 is 40 \times 20

132 can be rounded to 130 and 168 to 170.

Therefore, equivalent to 132 \div 168 is 130 \div 170.

17.3 can be rounded to 17.0 and 18.4 to 18.0.

Therefore, equivalent to 17.3 \times 18.4 is 17.0 \times 18.0.

999 can be rounded to 1000 and 208 to 210.

Therefore, equivalent to 999 \div 208 is 1000 \div 210.

998 can be rounded to 1000 and 211 to 210.

Therefore, equivalent to 998 \times 211 is 1000 \times 210.

The solution can be found in the attached image as well.

8 0
2 years ago
Helppp pleaseeeeee!!!
crimeas [40]
The correct answer is B C and D
5 0
3 years ago
What is the quotient. 4x^2+25x+12 divided by x+5
AveGali [126]

 4x2 - 25x + 27

 ——————————————

     x + 5  

3 0
2 years ago
Read 2 more answers
One positive number is one-fifth of another number. the difference of the two numbers is 84. find the numbers.
seraphim [82]
Answer:  The numbers are:  " 21 " and " 105 " .
___________________________________________________
Explanation:
___________________________________________________
Let "x" be the "one positive number:

Let "y" be the "[an]othyer number".

x = 1/5 (y)
___________________________________________________
Given that the difference of the two number is "84" ;  and that "x" is (1/5) of  "y" ;  we determine that "x" is smaller than "y".

So, y − x = 84 .

Add "x" to each side of this equation; to solve for "y" in terms of "x" ;

y − x + x = 84 + x  ;

 y = 84 + x ;
___________________________________________________
So, we have: 

 x = (1/5) y ;

and:  y = 84 + x  ;

Substitute "(1/5)y" for "x" ;  in  "y = 84 + x " ;  to solve for "y" ;

 y = 84 + [ (1/5)y ]

Subtract  " [ (1/5)y ] " from EACH SIDE of the equation ;

y − [ (1/5)y ] = 84 + [ (1/5)y ] −  [ (1/5)y ]  ;

to get:

  [ (4/5)y ] = 84 ;


       ↔    (4y) / 5 = 84  ;
      
        →  4y = 5 * 84  ;

      Divide EACH SIDE of the equation by "4" ; 
to isolate "y" on one side of the equation; and to solve for "y" ;

           4y / 4 = (5 * 84) / 4 ;

                 y =  5 * (84/4) = 5 * 21 = 105 .

   y = 105 .
___________________________________________________
Now, plug "105" for "y" into:
___________________________________________________
Either:
___________________________________________________
 x = (1/5) y ;

OR:

  y = 84 + x  ;
___________________________________________________
to solve for "x" ;
___________________________________________________
Let us do so in BOTH equations; to see if we get the same value for "x" ; which is a method to "double check" our answer ;
___________________________________________________
Start with:

x = (1/5)y 

    →  (1/5)*(105) = 105 / 5 = 21 ;  x = 21 ; 

___________________________________________________
So, x = 21;  y = 105 .
___________________________________________________
Now, let us see if this values hold true in the other equation:
___________________________________________________
y = 84 + x ;

105 = ? 84 + 21 ?
 
105 = ? 105 ? Yes!
___________________________________________________
The numbers are:  " 21 " and  "105 " .
___________________________________________________

6 0
3 years ago
If the total area of a dartboard is 30,000 mm2 and the area of the bull's-eye is 1000 mm2, what is the probability of getting a
AlladinOne [14]
Chances of hitting bull's-eye is \frac{1000}{30,000} that =\frac{1}{30} 1-:30= 0.03= 3% 
3 0
3 years ago
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