Plz help i beg uu. Will give branliest
2 answers:
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Answer:
5 units
Step-by-step explanation:
Let point O be the point of intersection of the kite diagonals.
|OF| = 2, |OH| = 5
|FH| = |OF| + |OH| = 2 + 5 = 7
FH and EG are the diagonals of the kite. Hence the area of thee kite is:
Area of kite EFGH = (FH * EG) / 2
Substituting:
35 = (7 * |EG|) / 2
|EG| * 7 = 70
|EG| = 10 units
The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units
x = |GO| = |EO| = 5 units
Answer:
Just took the test,
1)d, 3
2)a, -13
3) c, y=-2x+6
4) c, f'(a)=8a+2
Step-by-step explanation:
For #3, graph f(x)=-x^2+5. Next graph y=-2x+6. You will see that the added equation touches at exactly the point (1,4). this make it the tangent line equation
