The arched interior is comprised of a semi-circle and a rectangle. The diameter of the semi-circle is 20 inches. The length of t
1 answer:
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Answer:
Step-by-step explanation:
When a question asks for the "end behavior" of a function, they just want to know what happens if you trace the direction the function heads in for super low and super high values of x. In other words, they want to know what the graph is looking like as x heads for both positive and negative infinity. This might be sort of hard to visualize, so if you have a graphing utility, use it to double check yourself, but even without a graph, we can answer this question. For any function involving x^3, we know that the "parent graph" looks like the attached image. This is the "basic" look of any x^3 function; however, certain things can change the end behavior. You'll notice that in the attached graph, as x gets really really small, the function goes to negative infinity. As x gets very very big, the function goes to positive infinity.
Now, taking a look at your function, 2x^3 - x, things might change a little. Some things that change the end behavior of a graph include a negative coefficient for x^3, such as -x^3 or -5x^3. This would flip the graph over the y-axis, which would make the end behavior "swap", basically. Your function doesn't have a negative coefficient in front of x^3, so we're okay on that front, and it turns out your function has the same end behavior as the parent function, since no kind of reflection is occurring. I attached the graph of your function as well so you can see it, but what this means is that as x approaches infinity, or as x gets very big, your function also goes to infinity, and as x approaches negative infinity, or as x gets very small, your function goes to negative infinity.
A. The LCD is (n+1)(n-2)
B. n = -1, n = 2
C. n = negative 1 plus or minus the square root of 73, everything divided 2.
I’ve add a picture of the answers at the bottom in case I wasn’t clear on question C.
Hope that helps.
B. n = -1, n = 2
C. n = negative 1 plus or minus the square root of 73, everything divided 2.
I’ve add a picture of the answers at the bottom in case I wasn’t clear on question C.
Hope that helps.
9514 1404 393
Answer:
(a) x = (3 -ln(3))/5 ≈ 0.819722457734
(b) y = 10
Step-by-step explanation:
(a) Taking the natural log of both sides, we have ...
2x +1 = ln(3) +4 -3x
5x = ln(3) +3 . . . . . . . . add 3x-1
x = (ln(3) +3)/5 ≈ 0.820
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(b) Assuming "lg" means "log", the logarithm to base 10, we have ...
log(y -6) +log(y +15) = 2
(y -6)(y +15) = 100 . . . . . . . taking antilogs
y^2 -9x -190 = 0 . . . . . . . . eliminate parentheses, subtract 100
(y -19)(y +10) = 0 . . . . . . . . factor
The values of y that make these factors zero are -19 and 10. We know from the first term that (y-6) > 0, so y > 6. That means y = -19 is an extraneous solution.
The solution is ...
y = 10
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When solving equations using a graphing calculator, it often works well to define a function f(x) such that the solution is f(x) = 0, the x-intercept(s). That form is easily obtained by subtracting the right side of the equation from both sides of the equation. In part (a) here, that is ...
f(x) = e^(2x+1) -3e^(4-3x)
Answer:
third
Step-by-step explanation:
hope it helped!!!
