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bekas [8.4K]
3 years ago
14

The sum of two numbers is 0. Twice the smaller number subtracted from 3 times the larger number is 10. Let x represent th larger

number and y represent the smaller number Which equations represent this situation?

Mathematics
2 answers:
slava [35]3 years ago
5 0
The sum of two numbers is zero.
x + y = 0
y = -x

<span>Twice the smaller number subtracted from 3 times the larger number is 10.
Let x represent the larger number and y represent the smaller number.

Twice the smaller number: 2y

3 times the larger number: 3x

</span>Twice the smaller number subtracted from 3 times the larger number is 10. 
3x - 2y = 10

-2y = -3x + 10

y = 3/2 x - 5

The equations are:

y = -x
y = 3/2 x - 5

The answer is the first choice.
Anarel [89]3 years ago
5 0
-10 + 10 = 0

x+y = 0 y =(0-x)

3x-2y =10
3x-2(0-x)=10
3x-0 -2x=10
x=10

y=0-10 y = -10 FIRST ANSWER
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Diano4ka-milaya [45]

Answer:

B

Step-by-step explanation:

Starting with 7,000, after 0 years there will be no increase so you still have 7,000.

The fist year you increase by 5% of 7,000.

.05x7000=350

You have a 350 increase, add that to the original 7000 to find the actual population after 1 year (domain value 1).

After 1 year: 7350

For year 2 there is an increase of 5% again, only this time we find 5% of 7350 since that was the previous years population.

.05x7350=368

Add that to previous population.

368+7350=7718

At this point so far the yearly populations have been (7000, 7350, 7718)

Answer choice B is the only one to have this progression.

3 0
3 years ago
What is the value of this expression?<br><br> (37)0<br><br><br> 0<br><br> 37<br><br> 1<br><br> 307
Anni [7]

37^0 = 1

Any number to the 0 exponent is equal 1, except 0

Answer

1

5 0
3 years ago
Read 2 more answers
Using the equation x+3, write one equation that has one solution, one equation that has no solution, and one equation that has i
neonofarm [45]

9514 1404 393

Answer:

  • one: x+3 = 2x
  • none: x+3 = x
  • infinite: x+3 = x+3

Step-by-step explanation:

A linear equation with variable terms on opposite sides of the equal sign will have one solution when the coefficients of those variables are different.

  x+3 = 2x . . . . has one solution (x=3)

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There will be no solution if the variable terms on opposite sides of the equal sign have the same coefficient, but the constants are different. Such an equation can be reduced to 0 = 1, which cannot be made true by any value of the variable.

  x +3 = x . . . . has no solutions

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There will be an infinite number of solutions if the left side of the equal sign is the same as the right side. Every value of the variable will satisfy the equation.

  x +3 = x +3 . . . . has infinite solutions

8 0
2 years ago
The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}ordered position

Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

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3 years ago
Go on this one https://brainly.com/question/21200034<br><br> But other than that stay here.
Flura [38]

Answer:

ok thank you so much your the best

3 0
3 years ago
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