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olga_2 [115]
3 years ago
10

The Pythagorean theorem is a^2+b^2=c^2 Solve for b.

Mathematics
2 answers:
DaniilM [7]3 years ago
7 0

Answer:

B

Step-by-step explanation:

rodikova [14]3 years ago
3 0

Answer:

see explanation

Step-by-step explanation:

given

a² + b² = c² ( isolate b² by subtracting a² from both sides )

b² = c² - a² ( take the square root of both sides )

\sqrt{b^{2} } = ± \sqrt{c^2-a^2}

b = ± \sqrt{c^2-a^2}

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barxatty [35]

Answer: e = 17.8

Step-by-step explanation: 14.6 + 3.2

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3 years ago
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A dog chases a squirrel. The dog is originally 200 feet away from the squirrel. The dog's speed is 150 feet per minute. The squi
Dafna11 [192]

Answer:

15 seconds

Step-by-step explanation:

If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy:  the dog's rate is 150 and the squirrel's is 100.  The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know:  the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up.  Since we don't know what d is, we will just call it d!  Now it seems as though we have 2 unknowns which is a problem.  However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other.  Here's the dog's equation:

d = rt

d+200 = 150t

And the squirrel's:

d = 100t

If we solve both for t and set them equal to each other we have:

\frac{150}{d+200} =\frac{100}{d}

Now we can cross multiply to solve for d:

150d = 100d + 20,000 and

50d = 20,000

d = 400

But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took.  So sub that d value back into one of the equations we have solved for t and do the math:

t=\frac{100}{d}=\frac{100}{400} =\frac{1}{4}

That's 1/4 of a minute which is 15 seconds.

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Answer:

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3 years ago
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The intersection of the circle will result in either Tangent or Secant. Secant is a line intersecting a circle at two different point. This concept explains the relation between the circle and line intersecting it.

From the given diagram,

The \vec{GJ} and \vec{GK} intersects the circle at the point H and I.

∴ It can also represented that the \vec{JH} and \vec{KI} intersects at a point G.

From the given options \vec{KI} is the secant line that intersects the circle twice.

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What is the distance, d, between the points (3,5/2) and (5/3,1)?
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Answer:

The correct answer is \sqrt{145}/6

Step-by-step explanation:

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