890×(1+0.187÷12)^(12)−890×(1+0.125÷12)^(12)
=63.61 saved
Answer:
- 4 . 10
Step-by-step explanation:
Answer:
mBCD = 28°
Step-by-step explanation:
The angle mBFD inscribes the arc mBD, so we have that:
mBFD = mBD/2
76 = mBD/2
mBD = 152°
The angle mBOD is a central angle related to the arc mBD, so we have that:
mBOD = mBD = 152°
In the quadrilateral BODC, the sum of internal angles needs to be equal to 360° (property of all convex quadrilaterals). The angles mCBO and mCDO are right angles, because EDC and ABC are tangents to the circle.
So, we have that:
mBOD + mCDO + mBCD + mCBO = 360
152 + 90 + mBCD + 90 = 360
mBCD = 360 - 90 - 90 - 152
mBCD = 28°
Answer:
The answer in the procedure
Step-by-step explanation:
Let
A1 ------> the area of the first square painting
A2 ----> the area of the second square painting
D -----> the difference of the areas
we have


case 1) The area of the second square painting is greater than the area of the first square painting
The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting
D=A2-A1


case 2) The area of the first square painting is greater than the area of the second square painting
The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting
D=A1-A2

