<h2>
Answer:</h2>
A. It is a many-to-one function.
<h2>
Step-by-step explanation:</h2>
Hello! It will be a pleasure to help to figure out what's the correct answer to this problem. First of all, we have the following function:
When plotting this function, we get the red graph of the function shown below. So let's solve this as follows:
<h3>A. It is a many-to-one function.</h3>
True
A function is said to be many-to-one there are values of the dependent variable (y-values) that corresponds to more than one value of the independent variable (x-values). To test this, we need to use the Horizontal Line Test. So let's take the horizontal line 
, and you can see from the first figure below that is mapped onto 
. so this is a many-to-one function.
<h3>B. It is a one-to-one function.</h3><h3>False</h3>
Since this is a many-to-one function, it can't be a one-to-one function.
<h3>C. It is not a function.</h3>
False
Indeed, this is a function
<h3>D. It fails the vertical line test.</h3>
False
It passes the vertical line test because any vertical line can intersect the graph of the function at most once. An example of this is shown in the second figure below.
Answer:
<h3>
- 14</h3>
Step-by-step explanation:
{x - some integer}
2x - the first even integer
2x+2 - the second consecutive even integer
2x+4 - the third consecutive even integer (the largest)
2x + 2x+2 + 2x+4 - the sum of three consecutive even integers
2x + 2x+2 + 2x+4 = -48
6x + 6 = -48
÷3 ÷3
2x + 2 = -16
+2 +2
2x+4 = -14
Answer:
5-3x7^0
(5-3)7^0
2x1
2
Step-by-step explanation:
7^0 is 1
and do you want to chat
Answer:
20
Step-by-step explanation:
Because 21.50 x 20 =430
and 18 x 20 = 360
then you add 70 to 360 and you get 430
so Katrina would have to order 20 shirts from each company in order for the cost to be the same.
The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.
The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.
This leaves us with the second choice. Recall that
1 + tan²(x) = sec²(x)
and the derivative of tangent,
(tan(x))' = sec²(x)
Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then
y' = sec²(x³/3 + 2) • x²
and substituting y and y' into the ODE gives
sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))
x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)
which is an identity.
So the solution is y = tan(x³/3 + 2).
