now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12
now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8
now the last one on the far-right
has a height of 8, the Base is a Hexagon, with sides of 6
Answer:
Ignacio can make 5 clockwise rotations.
Step-by-step explanation:
Given that the Ignacio's legs are already at a height of 49.3cm and each rotation of the chair knob raises his legs another 4.8cm, we can set up an inequality to determine the number of rotations Ignacio could make without his legs touching the desk, which is at a height of 74.5cm:
4.8r + 49.3 < 74.5 where 'r' is equal to the number of rotations
The sum of the Ignacio's original leg height plus the amount of height increased from the rotations of the know must be less than 74.5 in order for his legs not to touch. Now, solve for 'r':
Subtract 49.3 from both sides: 4.8r + 49.3 - 49.3 < 74.5 - 49.3 or 4.8r < 25.2
Divide 4.8 from both sides: 4.8r/4.8 < 25.2/4.8 or r < 5.25
Since the number of rotations must be less than 5.25, he can make 5 complete rotations.
The answer is B.
Since ΔQRS ~ ΔXYZ, the value of tan(Q) is :
- ∠Q = ∠X
- tan(Q) = tan(X)
- tan(X) = 3/4
- tan(Q) = 3/4
2/5+3/8= 31/40 therefore he still has 9/40 to go.
to get this you multiply the two denominators together
5x8=40 which is your denominator
then you cross multiply the two numerators
2x8=16
3x5=15
to get the final
15/40+16/40=31/40
and you are left with 9/40
Answer:
-5
Step-by-step explanation:
Withdrawing means subtracting
