Answer:
Given: A triangle ABC and a line DE parallel to BC.
To prove: A line parallel to one side of a triangle divides the other two sides proportionally.
Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have
∠A=∠A (Common)
∠ADE=∠ABC (Corresponding angles)
Thus, by AA similarity, ΔABC is similar to ΔADE, therefore
AB/AD= AC/AE
⇒AD+DB/AD = AE+EC/AE
⇒1+DB/AD = 1+ EC/AE
⇒DB/AD = EC/AE
Therefore, a line parallel to one side of a triangle divides the other two sides proportionally.
⇒Therefore Proved
Hope this helps!!!
We have two equations that were solved by Nikki and Jonathon:
Equating the above two:
⇒ 1.3x + 1.6 = -2.7x + 3.2
⇒ 4x = 1.6
⇒ x = 0.4
Hence, substituting the value of x in one of the equations we get:
y = 1.3×0.4 + 1.6 = 2.12
So the solution is (0.4, 2.12)
Jonathon's solution was (0.4, 2.12) and Nikki's was (2.25, 0.5). Hence Jonathon gave the correct solution.
Answer:
Dilations produce similar figures because the image and pre-image will have congruent corresponding angles. The corresponding side lengths of the figures will be proportional based on the scale factor. The shape is preserved and the sides are enlarged or reduced by the scale factor
Step-by-step explanation:
Plug in n = 1 into the nth term formula
a(n) = 4n-1
a(1) = 4*1-1
a(1) = 3
So the first term is 3
The second term will be 7 because we add on 4 each time, as indicated by the slope of 4. This is also known as the common difference.
So the nth term is found by adding 4 to the (n-1)st term, in other words,
a(n) = a(n-1)+4
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In summary, the answer is
a1=3; an=an-1+4
which is choice B
