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AleksAgata [21]
3 years ago
14

Using substitution for systems of equations, solve, please show work. thank you!!

Mathematics
2 answers:
Rasek [7]3 years ago
5 0
That's your final answer. X=7 and Y=2

ratelena [41]3 years ago
4 0
X is equal to 7 and Y is equal to 2
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Suppose x = 5 is a solution to the equation 4x − 3(x + a) = 2. Find the value of a that makes the equation true. Group of answer
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Just replace all of the x’s with 5 and solve using order of operations. I will post what I got for my answer in the comments of the answer if you need it.
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Step-by-step explanation:

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A car with 15 gallons of gasoline drives 420 miles until the tank is empty. Write a linear equation that describes the amount of
storchak [24]

Answer:

y = -\dfrac{1}{28}x + 15

Step-by-step explanation:

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y = mx + 15

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5 0
3 years ago
Isaac records the following temperatures (in degrees Fahrenheit) at noon during one week:
Alecsey [184]

Answer:

first, we put them in order

78,[80],83,(85),86,[87],90

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Q3 = 87

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Hope this helps!

5 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
4 years ago
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