Answer:
5/24
Step-by-step explanation:
What are the angle measurements of 1,2,3,4, &5?
1 answer:
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Answer:
Step-by-step explanation:
Given the indefinite integral , using the substitute
x = 4 sec(θ)...1
The integral can be calculated as thus;
First let us diffrentiate the substitute function with respect to θ
dx/dθ = 4secθtanθ
dx = 4secθtanθdθ... 2
Substituting equation 1 and 2 into the integral function we will have;
Find the remaining solution in the attachment.
View the attached photo.
<span>The quadratic function, y = x2, has an x-intercept at the origin = TRUE
</span>
<span>The quadratic function, y = x2 + 3, has an x-intercept at the origin = FALSE
</span>
<span>From x = -2 to x = 0, the average rate of change for both functions is positive = FALSE
When viewing a graph, you start from the left and go to right and from left to right from -2 to 0 the graph is declining.
</span><span>From x = -2 to x = 0, the average rate of change for both functions is negative = TRUE
Both are declining if you read the graph from left to right.
</span><span>For the quadratic function, y = x2, the coordinate (2, 3) is a solution to the equation of the function. = FALSE
</span><span>
y = x^2
If we insert 2 into y = x^2 we get y = 4 but our point (2,3) has y = 3
</span>For the quadratic function, y = x2 + 3, the coordinate (2, 7) is a solution to the equation of the function. = TRUE
If we insert 2 for x, we see that y = x^2 + 3 = y = 2^2 + 3 = y = 7 and our y value in the point (2,7) is 7.
<span>The quadratic function, y = x2, has an x-intercept at the origin = TRUE
</span>
<span>The quadratic function, y = x2 + 3, has an x-intercept at the origin = FALSE
</span>
<span>From x = -2 to x = 0, the average rate of change for both functions is positive = FALSE
When viewing a graph, you start from the left and go to right and from left to right from -2 to 0 the graph is declining.
</span><span>From x = -2 to x = 0, the average rate of change for both functions is negative = TRUE
Both are declining if you read the graph from left to right.
</span><span>For the quadratic function, y = x2, the coordinate (2, 3) is a solution to the equation of the function. = FALSE
</span><span>
y = x^2
If we insert 2 into y = x^2 we get y = 4 but our point (2,3) has y = 3
</span>For the quadratic function, y = x2 + 3, the coordinate (2, 7) is a solution to the equation of the function. = TRUE
If we insert 2 for x, we see that y = x^2 + 3 = y = 2^2 + 3 = y = 7 and our y value in the point (2,7) is 7.