<em>c = 10</em>
Step-by-step explanation:
We can find out the missing side of a right triangle by using the Pythagorean theorem.
The Pythagorean theorem is...
We can even double check the first problem by plugging in everything into the theorem and solving, everything will come out correct. We can plug in the numbers from the second problem into the theorem and find c, also please note that the hypotenuse of a triangle will <em>always </em>be c. It doesn't matter which you put in for a or b, but since the problem gives us which one is a and which one is b, I'll just be plugging it in like that.
A lot of people will stop here and think that the answer is 100, but you need to find the square root of 100, since c is squared. The square root of 100 is 10, so...
<em><u>c = 10</u></em>
<h3>
Answer: 0.5</h3>
This is equivalent to the fraction 1/2
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Explanation:
The distance from A to B is 3 units. We can count out the spaces, or subtract the x coordinates of the two points and apply absolute value.
|A-B| = |-5-(-8)| = |-5+8| = |3| = 3
or
|B-A| = |-8-(-5)| = |-8+5| = |-3| = 3
We can say that segment AB is 3 units long.
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The distance from A' to B' is 1.5 units because...
|A'-B'| = |-2.5-(-4)| = |-2.5+4| = |1.5| = 1.5
or
|B'-A'| = |-4-(-2.5)| = |-4+2.5| = |-1.5| = 1.5
The absolute values ensure the distance is never negative.
We can say A'B' = 1.5
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Now divide the lengths of A'B' over AB to get the scale factor k
k = (A'B')/(AB)
k = (1.5)/(3)
k = 0.5
0.5 converts to the fraction 1/2.
The smaller rectangle A'B'C'D' has side lengths that are exactly 1/2 as long compared to the side lengths of ABCD.
Explain whether the points (−13,4), (−7,3), (−1,2), (5,1), (11,0), (17,−1) represent the set of all the solutions for the equati
ivann1987 [24]
Answer:
No, because the set of all solutions of y=−16x+116 is represented by the line of the equation.
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
The green is the starting place, when you move from a point on the green triangle to the blue one, you go 5 to the left and 3 down.
