Answer:
11. (x-2)/(x+8)
12. 9+t
13. h = 2A/b
Step-by-step explanation:
11. Since x is used to represent the number, 2 less than a number is x-2. 8 more than a number is x+8. The quotient of the first and second of these is the first one divided by the second one:
... (x -2)/(x +8) . . . . . parentheses are required unless the expression is typeset:
12. Apparently, we're to assume the salesman keeps on driving. If he had any sense (or a good union), he would stop after 9 hours. t hours added to 9 is ...
... 9 + t
13. Divide by the coefficient of h.
... h = A/((1/2)b) = 2A/b
Answer:
y = 3
Step-by-step explanation:
According to Euclidian theorem:
y^2 = 3*(12+3)
y^2 = 45 find the root of both sides
y = 3
Answer:
x= -11
Step-by-step explanation:
distribute the 3 and the 2: 3 - 9x = -8x + 14
add 9x from both sides and subtract 14 from both sides: -11 = x
x = -11
<span>12a^3b + 8a^2b^2 − 20ab^3
</span>12a^3b = 4ab(3a^2)
8a^2b^2 = 4ab(2ab)
20ab^3 = 4ab(5b^2)
GCF = 4ab
12a3b + 8a2b2 − 20ab3 = 4ab(3a^2 + 2ab - 5b^2)
Answer:
The equation does not have a real root in the interval ![\rm [0,1]](https://tex.z-dn.net/?f=%5Crm%20%5B0%2C1%5D)
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if 
is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval , which means to evaluate the equation in 0 and 1:
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval . I attached a plot of the equation in the interval ![\rm [-2,2]](https://tex.z-dn.net/?f=%5Crm%20%5B-2%2C2%5D)
where you can clearly observe how the graph does not cross the x-axis in the interval.
