If the perimeter is 16, the side length is 4 units.
To find the area of the shaded sections, you will use the fractional part that each section represents and multiply that fraction by the total area of the square.
Area of DEA is 1/4 of 16 square Units (4 x 4).
1/4 x 16 = 4 square units
Area of EFB is 1/8 of 16 square units.
1/8 x 16 = 2 square units
4 square units +2 square units equals 6 square units.
The area of the shaded region is 6 square units.
Answer:
7.2
Step-by-step explanation:
Given that in triangle LMN, LO is angle bisector of angle L.
LN =10 and LM =18
By angle bisector theorem for triangles we have
LN/LM = NO/MO
Substitute the values for known things and x for MO
We get
10/18 = 4/x
Or cross multiply to get
10x=72
x=7.2
So answer is 7.2
Answer: The r<span>-value for the linear function related to the ordered pairs is very close to zero, so it is not a good representation of the data. A quadratic model would better represent the data because there is a turning point within the data set. The data increases then decreases, which is what the graph of a quadratic does. </span>
Answer:
DA/dt = 75.27 cm²
Step-by-step explanation:
Cube Volume = V(c) = 683 cm³
DV(c) /dt = 824 cm³
V(c,x) = x³
Then
DV(c,x)/ dt = 3x² Dx/dt
( DV(c,x)/ dt )/ 3x² = Dx/dt (1)
Now as V(c,x) = x³ when V(c,x) = 683 cm³ x = ∛683
x = 8.806 ( from excel)
And by subtitution of this value in equation (1)
Dx/dt = ( DV(c,x)/ dt )/ 3x² ⇒ Dx/dt = 824 / 3*x²
Dx/dt = 824 /3*77.55
Dx/dt = 824/232,64
Dx/dt = 3,542
Then we got Dx/dt where x is cube edge. The area of the face is x² then
the rate of change of the suface area is
DA/dt = ( Dx/dt )²*6 ( 6 faces of a cube)
DA/dt = (3.542)²*6
DA/dt = 75.27 cm²
