In rhombus ABCD, the diagonal line AC and line BD intersect at E. If AE=5 and BE = 12, what is the length of line AB?
1 answer:
From a formula located here:
http://www.1728.org/quadltrl.htm
we see that
<span>4 • Side² = Long Diagonal² + Short Diagonal²
Long Diagonal = 24
Short Diagonal = 10
</span><span>4 • Side² = 24^2 + 10^2 </span>
<span>4 • Side² = 576 + 100
</span><span>4 • Side² = 676
</span><span><span>Side²</span> = 169
Side (or line AB) = 13
</span>
http://www.1728.org/quadltrl.htm
we see that
<span>4 • Side² = Long Diagonal² + Short Diagonal²
Long Diagonal = 24
Short Diagonal = 10
</span><span>4 • Side² = 24^2 + 10^2 </span>
<span>4 • Side² = 576 + 100
</span><span>4 • Side² = 676
</span><span><span>Side²</span> = 169
Side (or line AB) = 13
</span>
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Step-by-step explanation:
Divide two on both sides to get rid of it and the make the equation in the form y = mx + c
y = 3/2x + 5
since both lines are parallel they must have the same gradient which is 3/2
y = 3/2x + c
All you have to do now is to replace x and y with (2, -5) to find c
x = 2
y = -5
-5 = 3/2 × 2 + c
-5 = 3 + c
c = -5-3
c = -8
Answer:
The answer in the procedure
Step-by-step explanation:
The question does not present the graph, however it can be answered to help the student solve similar problems.
we know that
The equation of a vertical parabola into vertex form is equal to
where
a is a coefficient
(h,k) is the vertex
If the coefficient a is positive then the parabola open up and the vertex is a minimum
If the coefficient a is negative then the parabola open down and the vertex is a maximum
case A) we have
The vertex is the point (-4,-6)
a=3
therefore
The parabola open up, the vertex is a minimum
case B) we have
The vertex is the point (-4,-38)
a=3
therefore
The parabola open up, the vertex is a minimum
case C) we have
The vertex is the point (4,-6)
a=3
therefore
The parabola open up, the vertex is a minimum
case D) we have
The vertex is the point (4,-38)
a=3
therefore
The parabola open up, the vertex is a minimum