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andrezito [222]
3 years ago
7

Which line on the graph below has an undefined slope?

Mathematics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

R

Step-by-step explanation:

A vertical line has slope that is "undefined". Line R is the vertical line on your graph.

_____

The slope of a vertical line is "undefined" because of the way slope is defined. Slope is the vertical change divided by the horizontal change. When the line is vertical, there is no horizontal change, so the computation of slope involves division by zero. The result of division by zero is "undefined."

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If x = 6 and y=5, find y when x = 3
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Answer:

If x=6 and y=5 and we are to find y when x=3.

it will be 6=5,3=y....

5×3=6×y.

y=15/6=2.5 or 2 whole number 1/2.

This is the answer I hope it helps

8 0
3 years ago
1000973×21007÷710277-100=​
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Answer:

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Step-by-step explanation:

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Convert the linear equation 8x-2y=12 into function form.
kumpel [21]

Answer:

C

Step-by-step explanation:

get the y alone on one side: 2y=8x-12

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3 years ago
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The length of a rectangle is 6 yd longer than its width. if the perimeter of the rectangle is 48 yd , find its area.
Natasha_Volkova [10]
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3 0
3 years ago
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>

7 0
3 years ago
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