Given \qquad \overline{PQ}\perp\overline{PS} PQ ⊥ PS start overline, P, Q, end overline, \perp, start overline, P, S, end over
1 answer:
You might be interested in
ffn
the sum of their squares is 110, So
the sum of their cubes is 684, so
the product of the three integers is 210, so xyz= 210
the sum of any two products (xy+yz+zx) is 107
Now we plug in all the values in the identity
684 - 3(210) = (x+y+z)(110-107)
684 - 630 = (x+y+z)(3)
54 = 3(x+y+z)
Divide by 3 on both sides
18 = x+y+z
the value of the sum of three integers is 18
Answer:
Step-by-step explanation:
plug 1/4 into the formula for surface area and simplify to get 6/16, or 3/8.