If x = -7 then
<span>-3x+6y=-21
</span><span>-3(-7) +6y = -21
21 + 6y = -21
6y = -42
y = -7</span>
Answer: c) 0.75
Step-by-step explanation:
Given : The probability of choosing a black marble is P(Black)= 0.36.
The probability of choosing a black and then a white marble is P( Black and white) = 0.27.
Then by conditional probability ,
The probability of the second marble being white if the first marble chosen is black = 

Therefore , the probability of the second marble being white if the first marble chosen is black = 0.75
Answer:
+1 and -1
Step-by-step explanation:
The function in this problem is:

First of all, we have to define the domain of the function, which is the set of values of x for which the function is defined.
In order to find the domain, we have to require that the denominator is different from zero, so

which means:

So the domain is all values of x, except from 4 and -7.
Now we can solve the problem and find the zeros of the function. The zeros can be found by requiring that the numerator is equal to zero, so:

This is verified if either one of the two factors is equal to zero, therefore:

and

We see that both values are part of the domain, so they are acceptable values: so the zeros of the function are +1 and -1.
9×7=63
63÷2=31.5
31.5×4=126
7×7=49
126+49=175
175 is the answer
<u>the correct question is</u>
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.