Answer:
30
Step-by-step explanation:
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:
C
Step-by-step explanation:
It usually works best to use the polynomial with fewer terms as the multiplier. A row of partial products is written for each term of the multiplier, so the fewer terms will result in fewer rows of partial products.
In order to keep like terms together, it is preferable to allocate a separate column of the multiplication tableau to each power of the operands or product. This means we want to make note of the fact that the cubic multiplicand has a coefficient of 0 for its x^2 term.
The best setup is the one shown in the attachment.
Note the coordinates of each point: R(-4, 5), S(5, 1), T(2, -3).
The centroid is the point whose coordinates are the average of the coordinates of R, S, and T.
<em>x</em>-coordinate: (-4 + 5 + 2)/3 = 3/3 = 1
<em>y</em>-coordinate: (5 + 1 - 3)/3 = 3/3 = 1
So the centroid is (1, 1).
Search it up, you’ll get some answers. it maybe work