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weqwewe [10]
3 years ago
15

In addition to the car payment, Jackson and Bonnie estimate they will spend $200 on utilities, $140 on gasoline, $200 for grocer

ies, $55 for cell phone, and $45 for car insurance. What will be the total of the monthly bills?
Mathematics
2 answers:
salantis [7]3 years ago
7 0
The answer is 640.


Explanation: if you add them all together you can get the answer.
Studentka2010 [4]3 years ago
3 0
The total of the monthly bills will be $640.

(Add all of the bills together)
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Lines f and g look to be parallel to each other.

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6. Suppose a and b are integers and a^2 - 5b is even. Prove that b^2 - 5a is even.
cricket20 [7]

Answer:

See explanation below

Step-by-step explanation:

<u>First, let's see under which conditions a²- 5b is even.</u>

Case 1: a is even and b is even

If a is even then there exists a k >1 such that a = 2k

If b is even then there exists a j > 1 such that b = 2j

⇒a²- 5b = (2k)² - 5(2j) = 4k²-10j = 2( 2k² - 5j)

Therefore is a and b are even, then a²- 5b is even.

Case 2: a is even an b is odd

If a is even then there exists a k ≥ 1 such that a = 2k

If b is odd then there exists a j ≥ such that b = 2j - 1

⇒a²- 5b = (2k)² - 5(2j - 1) = 4k²- 10j + 5 = 4k²- 10j + 4 + 1 = 2 ( 2k² - 5j + 2) + 1

Therefore if a is even and b is odd a² - 5b is odd.

Case 3 : a is odd and b is odd

If a is odd then there exists a k ≥ 1 such that a = 2k  - 1

If b is odd then there exists a j ≥ such that b = 2j - 1

⇒a² - 5b = (2k - 1)² -5 (2j - 1) = 4k²- 4k +1 -10j  + 5 = 4k² - 4k -10j + 6 = 2 (2k² -2k -5j +3)

Therefore if a is odd and b is odd, a² - 5b is even.

Case 4: a is odd and b is even

If a is odd then there exists a k ≥ 1 such that a = 2k  - 1

If b is even then there exists a j ≥ such that b = 2j

⇒ a² - 5b = (2k -1)² - 5 (2j) = 4k² - 4k + 1 - 10j = 2( 2k²- 2k - 5j ) + 1

Therefore is a is odd and b is even, a² -5b is odd.

<u>So now we know that if a and b are integers and a² - 5b is even, then both a and b are odd or both are even.</u>

Now we're going to prove that b² - 5a is even for these both cases.

Case 1: If a² - 5b is even and a, b are even ⇒ b² - 5a is even

If a is even, then there exists a k≥1 such that a = 2k

If b is even, then there exists a j≥1 such that b = 2j

b² - 5a = (2j)² - 5(2k) = 4j² - 10k = 2 (2j² - 5k)

Therefore, b² - 5a is even

Case 2, If a² - 5b is even and a, b are odd ⇒ b² - 5a is even

If a is odd then there exists a k ≥ 1 such that a = 2k  - 1

If b is odd then there exists a j ≥ such that b = 2j - 1

b² - 5a = (2j - 1)² - 5(2k - 1) = 4j² - 2j + 1 -10k + 5 = 4j² -2j -10k + 6 = 2 (2j² - j - 5k +3)

Therefore, b² - 5a is even.

<u>Since we proved the only both cases possible, therefore we can conclude that if a and b are integers and a² - 5b is even, then b² - 5a is even.</u>

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No. She does not have enough money. She only has $251 and she needs at least $265 in order to buy the shoes.
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