Answer:
Hence Proved △ SPT ≅ △ UTQ
Step-by-step explanation:
Given: S, T, and U are the midpoints of Segment RP , segment PQ , and segment QR respectively of Δ PQR.
To prove: △ SPT ≅ △ UTQ
Proof:
∵ T is is the midpoint of PQ.
Hence PT = PQ ⇒equation 1
Now,Midpoint theorem is given below;
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
By, Midpoint theorem;
TS║QR
Also, ![TS = \frac{1}{2} QR](https://tex.z-dn.net/?f=TS%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20QR)
Hence, TS = QU (U is the midpoint QR) ⇒ equation 2
Also by Midpoint theorem;
TU║PR
Also, ![TU = \frac{1}{2} PR](https://tex.z-dn.net/?f=TU%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20PR)
Hence, TU = PS (S is the midpoint QR) ⇒ equation 3
Now in △SPT and △UTQ.
PT = PQ (from equation 1)
TS = QU (from equation 2)
PS = TU (from equation 3)
By S.S.S Congruence Property,
△ SPT ≅ △ UTQ ...... Hence Proved
Answer:
10 boxes
Step-by-step explanation:
Take the number of fossils and divide by the number per box to determine the number of boxes
60/6 = 10
She needs 10 boxes
Average= f(2)−f(0)/2−0
=62.5−250/ 2-0
<span>= −93.75</span>
So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
{1,3,7,9} as our answer.