<u>Answer:</u>
The degree of this polynomial is 3.
<u>Step-by-step explanation:</u>
The degree of a polynomial is the highest degree of its individual terms (also known as the monomials) which have a non-zero coefficient.
Here in this case, we have the following polynomial:
Each term has degree equal to the sum of the exponents on the variables. So for a term with two variables has the highest degree: 
1 for x and 2 for y: 1 + 2 = 3
Therefore, the degree of this polynomial is 3.
A quadratic has the form ax^2+bx+c
To factor a quadratic you want to find two values, j and k, which satisfy two conditions...
j*k=a*c and j+k=b
In this case ac=-140 and b=4.
The only values possible are j=14 and k=-10
so jk=-140 and j+k=4 which are the conditions we needed to satisfy...
Now you remove the linear, or "b term", from the original equation and replace it with jx and kx... and in this case you will have:
-2h^2-14h+10h+70 now you can factor out the greatest common factor from the first pair of terms and the second pair of terms...
-2h(h+7)+10(h+7) which is equal to
(-2h+10)(h+7) and you can factor the first term again if you wish...
-2(h-5)(h+7)
Answer:
1. B
3. C
4. B
5. What r the angles?
Step-by-step explanation:
Sry... I dont think I'm able to do #2 bc I'm only in Algebra I but there u go ;)
9514 1404 393
Answer:
- EF = DE = 44
- FG = DG = 36
- FH = DF = 31
Step-by-step explanation:
Since EH is the perpendicular bisector of DF, ∆DEF is isosceles and sides DE and EF have the same length.
DE = EF
(9x -1) = (7x +9)
2x = 10 . . . . . . . add 1-7x
x = 5 . . . . . . . . . divide by 2
__
Similarly, marked sides GD and GF are the same length, so ...
GD = GF
(10y -4) = (7y +8)
3y = 12 . . . . . . . . . . add 4-7y
y = 4 . . . . . . . . . divide by 3
__
Now, we have what we need to calculate the side lengths.
EF = 7x+9 = 7·5 +9 = 44
DE = 9x-1 = 9·4 -1 = 44
FG = 7y+8 = 7·4 +8 = 36
DG = 10y-4 = 10·4 -4 = 36
FH = 3x+4y = 3·5 +4·4 = 31
DF = FH = 31
