When graphing y = x2 – 14x + 24 and y = 20 – |x + 3|, which value would be the best minimum x- and minimum y-values for the viewing window so that all intercepts, minimums, maximums, and points of intersection are able to be seen as closely as possible? –50
–45
–25
–20

-25

Step-by-step explanation:

on edge, i took the test

 -45

It would be better if 30 were an option!

Step-by-step explanation:

At 25, all the important points will be in the graph-- barely. The vertex of the first equation is right at -25! So to be safe, it is better to go a bit larger.

If the viewing window is much larger, it is difficult to see small increments on the scale.

In the given triangle PRS, QS and RT are the altitudes. An altitude of a triangle is a line segment through a vertex and perpendicular to form a right angle with a line containing the base.

Area of triangle with an altitude = \frac{1}{2}\times base \times height

Area of the triangle with base PR and altitude QS = \frac{1}{2}\times PR \times QS

= \frac{1}{2}\times 8 \times 9 = \frac{72}{2}.

Area of the triangle with base PS and altitude RT = \frac{1}{2}\times PS \times RT

= \frac{1}{2}\times PS \times 7 = \frac{7PS}{2}.

by equating both the areas, we get,

\frac{72}{2}=\frac{7PS}{2}

PS=\frac{72}{7}= 10.3