Answer:
Yes
Step-by-step explanation:



<h3>Hope it is helpful....</h3>
Answer:
−11b^2+8b−4
Step-by-step explanation:
−4b2+6b−9−7b2+2b+5
Simplify by adding terms.
11b2+8b−4
I'd suggest you sketch this situation. If you did, you'd likely see right away that the shortest distance back to the starting point would be the hypotenuse of the right triangle whose legs are of lengths 5 miles and 12 miles.
Apply the Pythagorean Theorem to solve this.
If triangles PQR and STU are similar then PQ corresponds to ST and PR corresponds to SU. Therefore, PQ/ST=PR/SU
Considering that, PQ= 7-x, ST= 13-x, PR= x²+5 and SU= x² +20
therefore, (7-x)/(13-x)= (x²+5)/(x²+20)
cross multiplying,
7x² +140-x³+20x =13x²+65-x³-5x
combining the like terms,
6x² +15x -75=0
solving for x,
x = 5/2 or -5
The slope of the green line if the lines are perpendicular is -1/4
<h3>Perpendicular lines</h3>
For two lines two be perpendicular, the product of their slope must be -1. Let the slope of the red and green line be m1 and m2.
Given the following
Slope of red line = 4
According the definition
4m2 = -1
m2 = -1/4
Hence the slope of the green line if the lines are perpendicular is -1/4
Learn more on perpendicular lines here: brainly.com/question/1202004
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