D. 20 it's a 30-60-90 degree triangle so because the length of the hypotenuse is twice the length of the shortest leg
Sure.
Can you find a number that goes into both terms, 5k and 35 ?
How about 5 ?
(5k - 35) = 5 times (k - 7) .
Alright, so we have 1.3/0.0338. Since it's easier (in my opinion) to work with whole numbers, we can multiply the fraction by 10000/10000 to get 13000/338. With a bit of guess and check, we can see that
338*30=338*3*10
1 2 (what I carry is at the top)
338
x3
____
1114
Multiplying that by 10, I get 11140, which isn't enough. Trying 338*40, which is 338*4*10, we can add 338 to 338*3 to get 338*4 to get
2
1114
+338
____
1462
Multiplying that by 10, we get 14620, which is more than 13000 - something we don't want. Repeating this for 338*35 (which is 338*3.5*10, and 3.5 is 3*338+338/2)=11830 and which isn't enough, we then move on to something between 35 and 40 (the number doesn't matter), say 39. 338*39=338*3.9*10, and 338*3.9 is 338*3+338*9/10, and
338*39 results to 13182, which is more than 13000 , but only by a tiny bit, so we can try 38 using the same method, getting 12844, which is smaller, so we know it's between 38 and 39. Finding the difference between 13000 and 12844, we get 13000-12844=156 and the answer is therefore 38+156/338
Answer:
7.5
Step-by-step explanation:
Answer:
P ′(−10, −2), L ′(−5, −4), K ′(−15, 11)
Step-by-step explanation:
The translation vector, ⟨−6,3⟩, can be used to derive a rule for the translation of the coordinates: (x,y)→(x−6,y+3).
Apply the rule to translate each of the three preimage vertices to the image vertices.
P(−4,−5)→P'(−10,−2)
L(1,−7)→L'(−5,−4)
K(−9,8)→K'(−15,11)
Therefore, P'(−10,−2), L'(−5,−4), N'(−15,11) is the translation of the figure with the vertices P(−4,−5), L(1,−7), and K(−9,8), along the vector ⟨−6,3⟩.