How much is 3p-4p+r less than 4p+3q-5r​

3p−4q+r) is (p + 7q - 6r)(p+7q−6r) less than (4p+3q-5r)(4p+3q−5r) .

Let (3p-4q+r)(3p−4q+r) is 'x' less than (4p+3q-5r)(4p+3q−5r) . Then

3p-4q+r = 4p+3q-5r - x→3p−4q+r=4p+3q−5r−x

x= 4p+3q-5r - (3p-4q+r)→x=4p+3q−5r−(3p−4q+r)

x= 4p+3q-5r - 3p + 4q - r→x=4p+3q−5r−3p+4q−r

x= 4p- 3p +3q +4q - 5r - r→x=4p−3p+3q+4q−5r−r

x= p + 7q - 6r→x=p+7q−6r

Thus (3p-4q+r)(3p−4q+r) is (p + 7q - 6r)(p+7q−6r) less than (4p+3q-5r)(4p+3q−5r) .

5p+3q-6r

Step-by-step explanation:

4p+3q-5r -(3p-4p+r) = 4p+3q-5r-3p+4p-r = 5p+3q-6r

if u made a typo and the question is "How much is 3p-4q+r less than 4p+3q-5r", the answer is p+7q-6r​

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3