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Luda [366]
3 years ago
5

Which of these strategies would eliminate a variable in the system of equations?

Mathematics
2 answers:
valentinak56 [21]3 years ago
4 0
8x+8y=2
-) 8x+5y=1
——————
3y=1
y=1/3
8x+8(1/3)=2
24x+8=6
24x=-2
x=-1/12
(-1/12,1/3)
MrMuchimi3 years ago
3 0

Answer:

Step-by-step explanation:

8x + 8y = 2 -------(1)

8x + 5y = 1 -------(2)

subtract equation (2) from equation (1)

The 8x will be eliminated, leaving us with;

3y = 1

Divide bothside of the equation by 3

y = 1/3

substituting y = 1/3 in equation (1)

8x + 8y = 2

8x + 8(1/3) = 2

8x + 8/3 = 2

subtract 8/3 from both-side

8x + 8/3 -8/3 = 2 -8/3

8x = 2 - 8/3

8x = 6 - 8 / 3

8x = - 2/3

Divide both-side of the equation by 8

8x / 8 = -2/3 ÷ 8

x = -2/3 × 1/8

x = -1 / 12

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Rebecca and dan are biking in a national park for three days they rode 5 3/4 hours the first day and 6 4/5 hours the second day
AfilCa [17]

Answer:

Rebecca and Dan need to ride 7\frac{9}{20}\ hrs. on the third day in order to achieve goal of biking.

Step-by-step explanation:

Given:

Goal of Total number of hours of biking in park =20 hours.

Number of hours rode on first day = 5\frac34 \ hrs.

So we will convert mixed fraction into Improper fraction.

Now we can say that;

To Convert mixed fraction into Improper fraction multiply the whole number part by the fraction's denominator and then add that to the numerator,then write the result on top of the denominator.

5\frac34 \ hrs. can be Rewritten as \frac{23}{4}\ hrs

Number of hours rode on first day = \frac{23}{4}\ hrs

Also Given:

Number of hours rode on second day = 6\frac45 \ hrs

6\frac45 \ hrs can be Rewritten as \frac{34}{5}\ hrs.

Number of hours rode on second day = \frac{34}{5}\ hrs.

We need to find Number of hours she need to ride on third day in order to achieve the goal.

Solution:

Now we can say that;

Number of hours she need to ride on third day can be calculated by subtracting Number of hours rode on first day and Number of hours rode on second day from the Goal of Total number of hours of biking in park.

framing in equation form we get;

Number of hours she need to ride on third day = 20-\frac{23}{4}-\frac{34}{5}

Now we will use LCM to make the denominators common we get;

Number of hours she need to ride on third day = \frac{20\times20}{20}-\frac{23\times5}{4\times5}-\frac{34\times4}{5\times4}= \frac{400}{20}-\frac{115}{20}-\frac{136}{20}

Now denominators are common so we will solve the numerator we get;

Number of hours she need to ride on third day =\frac{400-115-136}{20}=\frac{149}{20}\ hrs \ \ Or \ \ 7\frac{9}{20}\ hrs.

Hence Rebecca and Dan need to ride 7\frac{9}{20}\ hrs. on the third day in order to achieve goal of biking.

3 0
3 years ago
Starting from the same place, Ryu walks due west and Samantha walks due east.
juin [17]

For Samantha, 2 tick marks east is 120 units east because 1 tick is 60 units.

For Ryu, 8 tick marks west is 480 units west because 1 tick is 60 units.

They are walking to form a line. Just add their distances from 0.

120 + 480 = 600.

--------------------------0----------------------------

Ryu: 480                         Samantha: 120

         480              +             120           =    600.

600 units, or feet.

5 0
3 years ago
How do you simplify this
kramer

Answer:

multiply the bottom and the top by route 2 minus route 3

this gives 4 route 2 minus route 18 over minus 1

this gives minus 4 route 2 add route 18

route 18 simplifies to 3 route 2

minus 4 route 2 add 3 route 2 gives minus route 2 as the answer

3 0
2 years ago
Samantha and her children went into a movie theater and will buy bags of popcorn and candies. Each bag of popcorn costs $6 and e
Juliette [100K]

Multiply the price of popcorn by the number of bags so 6b.

Multiply the price of candy by the number bought, so 3.25c

Add those together to get total :

6b + 3.25c

Now the most she can spend is 50 so set the equation to less than it equal to what she can spend:

6b + 3.25c <= 50

3 0
3 years ago
Given that
Ganezh [65]

Answer:

x^{2} -9x

Step-by-step explanation:

I will upload a picture with my work.

4 0
3 years ago
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