Consider the following number line
2 answers:
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X| 1 | 15 | 225 |
y| 4 | ? | 900 |
therefore
[tex]\dfrac{4}{1}=4;\ \dfrac{900}{225}=4;\ \dfrac{?}{15}=4\to?=60[\tex]
Answer: ? = 60
y| 4 | ? | 900 |
therefore
[tex]\dfrac{4}{1}=4;\ \dfrac{900}{225}=4;\ \dfrac{?}{15}=4\to?=60[\tex]
Answer: ? = 60
Answer:
x = ± - 3
Explanation:
I'm assuming you want the solutions to that equation, so here goes! (If not, please comment.)
(x-3)(x+9)=27
Let's FOIL this all out and expand. (Remember: First, Outer, Inner, Last.)
x^2 + 9x - 3x - 27
(first+ inner + outer + last)
x^2 + 9x - 3x - 27 = 27
Combine like terms, and add 27 to both sides.
x^2 + 6x - 27 + 27 = 27 + 27
x^2 + 6x = 54
Let's complete the square, because factoring doesn't work, and because it's good practice.
x^2 + 6x + ___ = 54 + ____
In the blank we will put b/2 ^2 = 6/2 ^2 = 3^2 = 9 to complete the square.
x^2 + 6x + 9 = 54 + 9
Now we've got a perfect square factor:
(x + 3)^2 = 63
sqrt(x+3)^2 =
x + 3 = ±
x = ± - 3