We are given an isosceles triangle.
An isosceles triangle has corresponding angles of corresponding sides same.
<em>Therefore, other angle is also of (3x+7) degrees.</em>
(5x+13) and (3x+7) makes a linear pair.
Therefore,
(5x+13) + (3x+7) = 180
5x +13 + 3x +7 = 180.
8x +20 = 180.
Subtracting 20 from both sides, we get
8x +20-20 = 180-20.
8x = 160
Dividing both sides by 8, we get
<h3>x = 20.</h3><h3>Therefore, correct option is 3rd option. </h3>
Answer:
no solution
Step-by-step explanation:
2x + 4y = 6......reduces to x + 2y = 3
3x = 12 - 6y....reduces to x = 4 - 2y...rearranged is x + 2y = 4
so now we have :
x + 2y = 3
x + 2y = 4
ok....I dont have to go any farther to know that this has no solution because ur equations have the same slope and different y int, this means ur lines are parallel and have no solution because they never cross each others path.
Answer:
134.73
Step-by-step explanation:
Solving for m:
243 = m + 108.27
243 - 108.27 = m + 108.27 - 108.27
134.73 = m
Answer:
For Lin's answer
Step-by-step explanation:
When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.
Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.
Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.
