Lower, initial, and at her starting weight
A) the x-values on the graph are [-2, 4].
all the x-values of the graph are the Domain of the function of the graph. they are found by projecting the graph to the x-axis. check the yellow line in the first picture.
b) all the possible outputs of g are [-1, 3].
all the possible outputs of g, are the Range of g. These are found by projecting the graph to the y-axis. check the red line in the second picture.
c) -1 , 0, 1, 2, and 3 are only a few values in the Range. The range is [-1, 3].
d) we will sketch a graph whose domain is [-2, 4] and range is [-1, 3]. check picture 3
Answer:
a) m∠3=113°
b) x=8 (but check below)
Step-by-step explanation:
a) we know m∠1=m∠5
6x-5=5x+7
x=12
we also know m∠1+m∠3=180°
plug x into the formula for m∠1
m∠1=67
m∠1+m∠3=180°
67+m∠3=180°
m∠3=113°
b) Again, we know since those lines are parallel that m∠4=m∠8
x²-4x=32
x²-4x-32=0
(x-8)(x+4)=0
x=8 or x=-4
i think (not 100% sure about this part lol) but x cannot be negative so it's 8
again i'm not sure but it is either 8 or 8,-4.
Answer:
100.
Step-by-step explanation:
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.