Answer:
8, 10
Step-by-step explanation:
Multiplying the equation by the product of the denominators, we have ...
40(x+2) +40x = 9x(x+2)
80x +80 = 9x^2 +18x . . . . eliminate parentheses
9x^2 -62x -80 = 0 . . . . . . put in standard form
9x^2 -72x +10x -80 = 0 . . . prepare to factor by grouping
9x(x -8) +10(x -8) = 0 . . . . . factor pairs of terms
(9x +10)(x -8) = 0
Solutions are ...
x = -10/9 and x = 8
The integers are 8 and 10.
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<em>Comment on the attachment</em>
I find a graphing calculator to be a nice tool for solving these. It finds zeros easily, so writing the equation so it equals zero is a useful first step.
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
d
Step-by-step explanation:
it goes with the correct statements in the sentences it is the right symbol and has the variable for the unknown amount of devices
Given:
The expressions are:
To find:
The value of given expression by using integer tiles.
Solution:
We have,
Here, both number are positive. When we add 6 and 3 positive integer tiles, we get 9 positive integer tiles as shown in the below figure. So,
Similarly,
Here, 6 is positive and -4 is negative. It means we have 6 positive integer tiles and 4 negative integer tiles.
When we cancel the positive and negative integer tiles, we get 2 positive integer tiles as shown in the below figure. So,
Here, 6 is positive and -6 is negative. It means we have 6 positive integer tiles and 6 negative integer tiles.
When we cancel the positive and negative integer tiles, we get 0 integer tiles as shown in the below figure. So,
Therefore, 
.
