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Svet_ta [14]
3 years ago
10

Determine whether the graphs of the given equations are parallel, perpendicular, or neither. Explain.

Mathematics
1 answer:
mr_godi [17]3 years ago
3 0

Answer:

They are neither parallel nor perpendicular.

it isn't parallel because they cross

it's not perpendicular because they don't cross at a 90 degree angle

Step-by-step explanation:

Hope this helps you and btw u should download desmos it's an amazing calculator that graphs these things for you!!! hope u have a wonderful day and stay safe and healthy:)))

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The number line represents Allie’s scores in the first two rounds of a card game.
olganol [36]

Answer:

Allie gains 4, she loses 4, she has 0 points.

Step-by-step explanation:

The arrow shows it. Round 1 goes 0 to 4. Round 2 goes 4 to 0.

7 0
3 years ago
I am stuck on this question of my maths test.
V125BC [204]

Answer:

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96 + y = 180

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the values of x = 29 and y = 84

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A conical tank is leaking water at the rate of 75 cubic inc/min. At the same time water is being pumped into the tank at a const
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Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
What is a counterexample of the statement “all square roots are irrational”
skad [1K]

Rather than trying to guess and check, we can actually construct a counterexample to the statement.

So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:

\sqrt{(\frac{3}{2})^2}=\\\sqrt{\frac{9}{4}}=\\\frac{3}{2}

So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let \frac{a}{b} be a rational number. That would mean, \frac{a^2}{b^2}, would be a/b squared. Taking the square root of it yields:

\sqrt{\frac{a^2}{b^2}}}=\\ \frac{\sqrt{a^2}}{\sqrt{b^2}}=\\ \frac{a}{b}

So our stronger statement is proven, and we know that the original claim is decisively false.

3 0
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