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Maurinko [17]
3 years ago
14

Is there an ordered pair that is a solution to BOTH of

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
6 0

Answer: The ordered pair is (2,3)

Step-by-step explanation:

To solve this problem, we can use substitution. With substitution, we substitute the y in each problem with the equations. In this case, x + 1 can substitute the y in y = -x + 5. Here is what the problem looks like now:

x + 1 = -x + 5

The reason we used substitution is because we assume that these linear equations are equal. We must prove it, though. We can use algebra.

Move the variables to one side and the numerical values to another:

x + x = 5 - 1.

Simplify.

2x = 4

Simplify.

x = 2.

Now, we got our x value, we can substitute it into our problem:

y = 2 + 1

y + -2 + 5

2 + 1 = - 2 + 5 = 3

Our x is 2, and our y is 3. The ordered pair that is a solution is (2,3).

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Find the variable h+ 1/2 = 3 1/4
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Step-by-step explanation:

2 3/4+1/2=

you can convert 1/2 into 2/4 so they have like denominators

2 3/4+2/4= 2 5/4 which is also equal to 3 1/4

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Write a formula for the general term or nth term for the sequence. Then find the indicated term. five halves comma five fourths
nadya68 [22]
Sequence: 5/2, 5/4, 5/8, 5/16
a8=?

a1=5/2
a2=5/4
a3=5/8
a4=5/16

a2/a1=(5/4)/(5/2)=(5/4)*(2/5)=(5*2)/(4*5)=2/4=1/2
a3/a2=(5/8)/(5/4)=(5/8)*(4/5)=(5*4)/(8*5)=4/8=1/2
a4/a3=(5/16)/(5/8)=(5/16)*(8/5)=(5*8)/(16*5)=8/16=1/2
Ratio: r=a2/a1=a3/a2=a4/a3→r=1/2

an=a1*r^(n-1)
a1=5/2, r=1/2
an=(5/2)*(1/2)^(n-1)
an=(5/2)*[1^(n-1)/2^(n-1)]
an=(5/2)*[1/2^(n-1)]
an=(5*1)/[2*2^(n-1)]
an=5/2^(1+n-1)
an=5/2^n

n=8→a8=5/2^8
a8=5/256

Answers:
The formula for the general term or nth term for the sequence is an=5/2^n
a8=5/256
6 0
3 years ago
How many solutions would there be to make an equation true?
mariarad [96]

Answer:

Step-by-step explanation:

Only one

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