Algebra 1. Question 13

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Hello!

The domain is the set of all possible x-values which will make the function true, and will output real y-values. Filled in points make the function true, and is written as ≤, but open circles, will make the function false and is written as just a less than symbol (<).

In this case, the ordered pair (-5, -2) is part of our domain, while the ordered pair (4, 1) is not because it is an open circle.

Therefore, the domain of the function is -5 ≤ x < 4, [-5, 4), {x | -5 ≤ x < 4}.

The range is the y-values of the function. In this case, the point (1, -4) is part of our range is also the minimum y-value, -4. The maximum y-value is the vertex of the parabola (the curve), and is the ordered pair, (-2, 3).

Therefore, the range of the function is -4 ≤ y ≤ 3, [-4, 3], {y | -4 ≤ y ≤ 3}.

Note: I wrote the domain and range in three ways because it doesn't specify which way to write it.

You might be interested in

Anyone know how to do this?

Novay_Z [31]

When you add 2pi to the angle, the amount is still the same. Therefore, you can make -17pi/4 positive by adding 6pi to it.

6pi + (-17pi/4)
= 24pi/4 - 17pi/4
= 7pi/4

8 0

3 years ago

Evaluate the integral by making an appropriate change of variables.

VARVARA [1.3K]

By inspecting the integrand, the "obvious" choice for substitution would be

u = y + x

v = y - x

Solving for x and y, we would have

x = (u - v)/2

y = (u + v)/2

in which case the Jacobian and its determinant are

$J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12$