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The length of AB to the nearest hundredth

The length of AB to the nearest hundredth-example-1
User LoganHenderson
by
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2 Answers

4 votes
4 votes

Answer: 7.28

Explanation:

This answer is solved by using the Pythagorean Theorem in a clever way. As we know, the Pythagorean Theorem states that:
a^2 + b^2 = c^2

Rewriting this, we find out that it:
c = √(a^2 + b^2)

We are using the value C to be the length of AB.

Now that we know how we are solving for c, we can now find values for a and b.

What we want to do, is find sides that will complete a triangle with AB. These sides are measures as:
x_2 - x_1 (where
x_2 is the x value of coordinate B, and
x_1 is the x-value of coordinate A) and
y_2 - y_1 (where y is similar to x).

Now that we have our sides, we can plug them in to the theorem we rewrote before:


c = √((x_2 - x_1)^2 + (y_2 - y_1)^2) = √((-3 - (-5))^2 + (3 - (-4))^2)

We can then begin to solve for AB:


AB = √((2)^2 + (7)^2) = √(4 + 49) = √(53) = 7.28

I hope this helps. If you need anything explained more please let me know.

User Aaron Gray
by
3.5k points
5 votes
5 votes

Answer:

8

Explanation:

you count the blocks starting from a to b

User Jason Yeo
by
3.5k points