There is a square and a triangle
the area of a square is
A = b * h
A = 2*2
A = 4
the area of the triangle is
A = (b*h)/2
A = (2 *2)/2
A = 4/2
A = 2
Now we add them up to find the total area
4 + 2 = 6
The shape has an area of 6 units^2
Problem 1
<h3>Answer: B. M<3 would need to double.</h3>
Explanation: This is because angles 3 and 6 are congruent corresponding angles. Corresponding angles are congruent whenever we have parallel lines like this. If they weren't congruent, then the lines wouldn't be parallel. We would need to double angle 3 to keep up with angle 6.
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Problem 2
<h3>Answer: D. none of these sides are parallel</h3>
Explanation: We have angles A and C that are same side interior angles, but they add to A+C = 72+72 = 144, which is not 180. The same side interior angles must add to 180 degrees for parallel lines to form. This shows AB is not parallel to CD.
A similar situation happens with angles B and D, since B+D = 108+108 = 216. This also shows AB is not parallel to CD. We can rule out choices A and C.
Choice B is false because AD is a diagonal along with BC. The diagonals of any quadrilateral are never parallel as they intersect inside the quadrilateral. Parallel lines never intersect.
The only thing left is choice D. We would say that AC || BD, since A+B = 72+108 = 180 and C+D = 72+108 = 180, but this isn't listed as an answer choice.
<span>P = 2l + 2w
The given values fit the given formula</span>
The given values are 134
Let L = 5 - w
Solution:
P = 2l + 2w
134 = 2(5-w) + 2w 134 = 10-2w + 2w 134 – 10 = 4w 124 = 4w
31 = w
<span>26 = l</span>
The ordered pair (2,7), is not a solution to the equation
9x+3 = -9 and 9x+3 = +9
x=-12/9
x=6/9
