<span>jika xy = 0 , kemudian menganggap x dan y = 0
faktor 2x^2-9+7
(2x-2)(x-7)=0
2x-2=0
2x=2
x=1
x-7=0
x=7
</span><span>jika 1 = x1 dan 7 = x2 maka jawabannya adalah 1^2+7^2-4(1)(7)=22
</span>
jika 7 = x1 dan 1 = x2 maka jawabannya adalah
7^2+7^2-4(7)(1)=22
<span>jawabannya adalah 22</span>
7.
(2b^2+7b^2+b)+(2b^2-4b-12)
(9b^2+b)+(2b^2-4b-12)
9b^2+b+2b^2-4b-12
11b^2+b-4b-12
11b^2-3b-12
8.
(7g^3+4g-1)+(2g^2-6g+2)
7g^3+4g-1+2g^2-6g+2
7g^3-2g-1+2g^2+2
7g^3-2g+1+2g^2
7g^3+2g^2-2g+1
Hope this helps!
The answer is D.
Explanation- 28/7=4
19-4 =15
This =2.4 I hope that this helps you
It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}