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Valentin [98]
3 years ago
7

the linear combination method is applied to a system of equations as shown.4(.25x .5y = 3.75) → x 2y = 15 (4x – 8y = 12) → x – 2

y = 3 2x = 18what is the solution of the system of equations?(1,2)(3,9)(5,5)(9,3)
Mathematics
1 answer:
gtnhenbr [62]3 years ago
4 0
If we use the last equation:
2 x = 18
x = 18 : 2
x = 9
Then we will put it in another equation:
 x - 2 y = 3
 9 - 2 y = 3
 - 2 y = 3 - 9
 - 2 y = -6
  y = -9 : ( -3 ) 
  y = 3.
   The solution is : D ) ( 9, 3 )
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Need help on 15 and 16 PLEASEE!!!
melisa1 [442]

Answer:

Part 15) The next three terms are 5/8,3/4 and 7/8

Part 16) The 37th term is -35.5

Step-by-step explanation:

we know that

In an <u>Arithmetic Sequence</u> the difference between one term and the next is a constant and this constant is called the common difference (d)

Part 15) Find the next three terms of the arithmetic sequence

\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},...

so

a1=1/8\\ a2=1/4\\ a3=3/8\\ a4=1/2\\ a2-a1=1/4-1/8=1/8\\ a3-a2=3/8-1/4=1/8\\ a4=a3=1/2-3/8=1/8

The common difference is equal to

d=1/8

<u><em>Find the next three terms of the arithmetic sequence</em></u>

<u><em>Find a5</em></u>

a5=a4+d ----->a5=1/2+1/8=5/8

<u><em>Find a6</em></u>

a6=a5+d ----->a6=5/8+1/8=3/4

<u><em>Find a7</em></u>

a7=a6+d ----->a7=3/4+1/8=7/8

therefore

The next three terms are

5/8,3/4 and 7/8

Part 16) what is the 37th term of the arithmetic sequence

4.1,3,1.9,0.8,...?

so

a1=4.1\\ a2=3\\ a3=1.9\\ a4=0.8

a2-a1=3-4.1=-1.1\\ a3-a2=1.9-3=-1.1\\ a4=a3=0.8-1.9=-1.1

The common difference is equal to

d=-1.1

We can write an Arithmetic Sequence as a rule

an=a1+d(n-1)

substitute the values

an=4.1-1.1(n-1)

For n=37

substitute

a37=4.1-1.1(37-1)

a37=4.1-1.1(36)

a37=-35.5

4 0
2 years ago
I only need help on number 9. Find the value of Y. ​
natima [27]
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3 0
2 years ago
I’ll be very grateful for anyone’s help I added a photo of the question
Irina18 [472]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A computer training institute has 625 students that are paying a course fee of $400. Their research shows that for every $20 red
Anastasy [175]
1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is: 

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have, 

a fee of $380 means that n=1, thus 

250(n+50)(20-n)=250(1+50)(20-1)=242,250   ($)


a fee of $320 means that n=4, thus

 250(n+50)(20-n)=250(4+50)(20-4)=216,000    ($)


the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
6 0
3 years ago
-3x-2y=-1<br> 4x+6y=12<br><br> what’s all the steps and answer to this equation
abruzzese [7]

Answer:

x = -9/5

y = 16/5

Step-by-step explanation:

-3x-2y=-1            first you need to multiply the numerator by 3 so when adding

4x+6y=12           the y's cancel and you get a single variable

3(-3x-2y=-1)     —>  -9x - 6y = -3     then you add the like terms

 4x+6y=12      —>   4x + 6y = 12

(-9x+4x) + (6y - 6y) = (-3 + 12)

-5x = 9  then you divide by -5

x = -9/5

then you plug this in to either equation

4(-9/5) + 6y = 12

-36/5 + 6y = 12   then you add 36/5 to both sides

6y = 96/5  then you divide by 6

y = 16/5

6 0
3 years ago
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