Evaluate x(–y + z) for x = –1, y = 2, and z = 5. 7 –7 3 –3

1 answer:

8 0

X(-y+z)

When x is -1, y is 2, and z is 5, replace these letters with the numbers associated with them, you get

=-1(-2+5)
=-1(-2)+(-1)(5)
=2-5
=-3

You might be interested in

Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.

user100 [1]

When x=-1: $\quad (-1)y=-4\qquad\to\qquad y=4\)$

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

$x'y+xy'=0$

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4

and solve for x'.

Hope that helps!

4 0

3 years ago

The equation f(x) = 3x2 − 24x + 8 represents a parabola. What is the vertex of the parabola?

s2008m [1.1K]

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{3}x^2\stackrel{\stackrel{b}{\downarrow }}{-24}x\stackrel{\stackrel{c}{\downarrow }}{+8} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{-24}{2(3)}~~,~~8-\cfrac{(-24)^2}{4(3)} \right)\implies (4~~,~~8-48)\implies (4~~,~-40)$