W² - 49 = 0
w² - 7² = -
(w-7)(w+7)=0
w -7=0 , or w + 7 =0
w=-7, w=7
Answer: -7, 7.
The variable m should equal -5
Question:
Morgan is playing a board game that requires three standard dice to be thrown at one time. Each die has six sides, with one of the numbers 1 through 6 on each side. She has one throw of the dice left, and she needs a 17 to win the game. What is the probability that Morgan wins the game (order matters)?
Answer:
1/72
Step-by-step explanation:
<em>Morgan can roll a 17 in 3 different ways. The first way is if the first die comes up 5, the second die comes up 6, and the third die comes up 6. The second way is if the first die comes up 6, the second die comes up 5, and the third die comes up 6. The third way is if the first die comes up 6, the second die comes up 6, and the third die comes up 5. For each way, the probability of it occurring is 1/6 x 1/6 x 1/6 = 1/216. Therefore, since there are 3 different ways to roll a 17, the probability that Morgan rolls a 17 and wins the game is 1/216 + 1/216 + 1/216 = 3/216 = 1/72</em>
<em>I had this same question on my test!</em>
<em>Hope this helped! Good Luck! ~LILZ</em>
You are given two points that could be plotted on a graph (1,-5) and (4,1). The formula for slope is y2-y1/x2 -x1
When plotted into the points it would look like 1-(-5)/4-1= 6/3
6/3=2. So, the slope for these two points is 2.
To factor this fraction, you have be be aware of two special factoring formula:
a^3<span> + </span>b^3<span> = (</span>a<span> + </span>b)(a^2<span> – </span>ab<span> + </span>b^2<span>)
</span><span>(a+b)³ = a³ + 3a²b + 3ab² + b³
You can see the top part in this case is (x+y)^3, and the bottom (denominator) can be factor into (x+y)(x^2-xy+y^2)
we can cancel (x+y), so what we have left is (x+y)^2/(x^2-xy+y^2)
or (x^2+2xy+y^2)/(x^2-xy+y^2)
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