Answer:
Let two consecutive multiples of 3 be x and (x+3)
A/q,
x * (x+3) = 648
➡ x² + 3x = 648
➡ x² + 3x -648 = 0
➡ x² + 27x - 24x -648 = 0
➡ x ( x + 27 ) -24 ( x +27)
➡ ( x - 24) ( x + 27)
➡ x = 24 and x = -27
so, we take x = 24.
Required multiples of 3
➡ x = 24
➡ x +3 = 24+3 = 27.
Answer:
14
Step-by-step explanation:
-3u = 4+5
-3u = 9
u = -3
7u + 7
7(u +1) = 7(-3+1)
7(-2)
14
C.25 because 11-6 is five 13-8 is five and 15-10 is five so 25-20 is five
Answer:
20
Step-by-step explanation:
50 - 20 = 30
30 ÷ 1.50 = 20
There are two methods to solve this problem. One of them is the typical method and the other one is the short-cut. I will explain the shortcut
Short-cut:
This method is quick and easy, and works with ALL triangles, which is why I suggest you use this. You make proportions. Since the triangles are similar, we can use CPCTC. Thereafter,

Now, you just need to identify the terms, substitute, and solve for the unknown, which would be k.

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So we found k, but it is asking for BE. Therefore, plug in the value of k, which is 11, into the expression for BE, which is k-7.
BE=K-7
k=11
BE=11-7
BE=4
Your answer is 4.