EASY POINTS HERE FOLKS COME HELP ME OUT
2 answers:
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A. If you plot the points in a graph, it would look like that shown in the picture attached. If we use linear regression, the correlation is very poor. The coefficient of correlation (r2) is only 0.0017. There is no linear relationship between time and velocity.
B. The slope of the graph is equal to y2-y1/x2-x1, In this case, it would specifically be v2-v1/t2-t1
Slope = 0.8-0.2/20-10 = 0.06 miles/s^2
The slope represents the acceleration at time 10 to 20 minutes.
C. The table in the graph shows causation rather than correlation. The points in the data occur in a sequential manner.
B. The slope of the graph is equal to y2-y1/x2-x1, In this case, it would specifically be v2-v1/t2-t1
Slope = 0.8-0.2/20-10 = 0.06 miles/s^2
The slope represents the acceleration at time 10 to 20 minutes.
C. The table in the graph shows causation rather than correlation. The points in the data occur in a sequential manner.
Answer:
Step-by-step explanation:
1) Statements Reasons
ABCD is a trapezoid Given
AD ║ BC bases of trapezoid are parallel
∠AED = ∠CEB vertically opposite angles are equal
∠EBC = ∠ADE alternate interior angles are equal and BD is transversal
∠ECB = ∠EAD alternate interior angles are equal and AC is transversal
ΔAED ~ ΔCEB ∠ECB = ∠EAD & ∠AED = ∠BEC & ∠EBC = ∠ADE
all angles are same and their shape is same, so similar
2) hope you can write 2nd one as above two column proof
T is the mid point of QR
U is the mid point of QS
V is the mid point of RS
Definition: The segment of line joining the middle points of two sides of a triangle is called middle segment.
Inference: A middle segment of a triangle is parallel to the third side and its lengths is half the third side's.
so here TU ║ RV and TU = RV
UV ║ TR and UV = TR
so TUVR becomes a parallelogram
∠TUV = ∠TRV --------> condition 1
same goes with parallelogram TQUV
∠TQU = ∠UVT ----------> condition 2
same goes with parallelogram TUSV
∠UTV = ∠USV ------------> condition 3
from the above 3 conditions we can say
ΔQRS ~ ΔVUT
3) see the below figure for graph
in both triangles ∠B is common
AC and TS are parallel lines
BC and BA are transversals
∠C ∠S are equal --> corresponding angles
∠A ∠T are equal ---> corresponding angles
so ΔABC & ΔTBS are similar
