Write 27 as a power and as repeated multiplication
2 answers:
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Let
A------> <span>(1, 4)
B------> (6, 4)
C-----> (6,1)
we know that
In </span><span>a right triangle there are two legs and one hypotenuse
</span><span>
step 1
find the distance AB
d=</span>√[(y2-y1)²+(x2-x1)²]------> d=√[(4-4)²+(6-1)²]-----> d=√25----> 5
AB=5 units
step 2
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> d=√[(1-4)²+(6-6)²]-----> d=√9----> 3
BC=3 units
step 3
find the distance AC
d=√[(y2-y1)²+(x2-x1)²]------> d=√[(1-4)²+(6-1)²]-----> d=√34----> 5.83
AC=5.83 units
therefore
the two legs are
AB=5 units
BC=3 units
the hypotenuse is
AC=5.83 units
the answer is
<span>the length of the longer leg is AB=5 units
</span>
see the attached figure
A------> <span>(1, 4)
B------> (6, 4)
C-----> (6,1)
we know that
In </span><span>a right triangle there are two legs and one hypotenuse
</span><span>
step 1
find the distance AB
d=</span>√[(y2-y1)²+(x2-x1)²]------> d=√[(4-4)²+(6-1)²]-----> d=√25----> 5
AB=5 units
step 2
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> d=√[(1-4)²+(6-6)²]-----> d=√9----> 3
BC=3 units
step 3
find the distance AC
d=√[(y2-y1)²+(x2-x1)²]------> d=√[(1-4)²+(6-1)²]-----> d=√34----> 5.83
AC=5.83 units
therefore
the two legs are
AB=5 units
BC=3 units
the hypotenuse is
AC=5.83 units
the answer is
<span>the length of the longer leg is AB=5 units
</span>
see the attached figure
bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.