Answer:
The volume of cone z has the same base area as cylinder y but its height is 3 times the height of cylinder y have the same volume.
Step-by-step explanation:
The volume of a cylinder having base area 'A' and height 'H' is given by V = AH, and the volume of a cone having the same base area as the cylinder i.e. A and height 3 times the height of the cylinder i.e. 3H is given by 
.
So, V = V'
Hence, the volume of cone z has the same base area as cylinder y but its height is 3 times the height of cylinder y have the same volume. (Answer)
Answer:
<em>total</em><em>=</em><em>n÷m</em><em> </em><em>$</em><em>1</em><em>6</em><em>5</em><em>÷</em><em>1</em><em>5</em><em>,</em><em>$</em><em>2</em><em>3</em><em>1</em><em>÷</em><em>2</em><em>1</em><em> </em><em>and</em><em> </em><em>$</em><em>2</em><em>7</em><em>5</em><em>÷</em><em>2</em><em>5</em>
Answer:
Option 3x+5y=29 and -3x-12y=-48 is the system of equations equivalent to the given system of equations 3x+5y=29 and x+4y=16
Step-by-step explanation:
Given system of equations is 3x+5y=29 and x+4y=16
To find the equivalent system of equations to the given system of equations :
Option 3x+5y=29 and -3x-12y=-48 is the system of equations represents the given system of equations.
Because we can write the given equations as below
3x+5y=29 and x+4y=16
x+4y=16 rewritting as below
When multiply the above equation into (-3) we get
as same as the equation x+4y=16 so we can say that they are equivalent
Therefore Option 3x+5y=29 and -3x-12y=-48 is the system of equations represents the given system of equations
Therefore 3x+5y=29 and -3x-12y=-48 is the system of equations equivalent to the given system of equations 3x+5y=29 and x+4y=16
Answer:
m∠DEA = 62°
m∠ADB (arc) = 194°
Angle ∠ADB = 21°.
Step-by-step explanation:
The given information are;
, m CB (arc) = 62°, m∠DAB (arc) = 104°
arc∠BCD = 360° - 104° = 256°
m DC (arc) = arc∠BCD - arc CB = 256° - 62° (Sum of angles)
Therefore DC (arc) = 194°
m DA ≅ m CB = 62° (Parallel lines congruent arc theorem. Arc between two parallel lines)
m∠DEA = 1/2×(arc DA + arc CB) = 1/2×(62° + 62°) =62°
m∠DEA = 62°
Arc AB = m∠DAB (arc) - m DA = 104° - 62° = 42°
m∠ADB (arc) = 360 - m∠DAB (arc) - m CB (arc) (Sum of angles around a circle or point)
∴ m∠ADB (arc) = 360 - 104 - 62 = 194°
m∠ADB (arc) = 194°
Angle ADB = subtended by arc AB = ∴1/2×arc AB
Angle ∠ADB = 42/2 = 21°.
Angle = 21°
