There are two triangular faces and three rectangular faces.
The triangular faces are congruent, so they have the same area. We calculate one and we'll know both.
The three rectangles are different.
Triangular face (each)
Although you can hardly see in the colored figure, there is a small symbol on the right side showing that the 4 km side and the 3 km side of the triangles are perpendicular. That means those sides are the base and the height.
A = (1/2)bh = (1/2)(3 km)(4 km) = 6 km^2
Rectangular faces:
A = LW = 7 km * 3 km = 21 km^2
A = LW = 7 km * 5 km = 35 km^2
A = LW = 7 km * 4 km = 28 km^2
total surface area = 2 * 6 km^2 + 21 km^2 + 35 km^2 + 28 km^2
total surface area = 96 km^2
I could be very wrong, but I believe the answer is C
Let x be for unknown number. Let x be first number .
X+1 To be second number .x+2be third number. X+(x+1)+(x+2)=312.
Simplify to get 3x+3=312 . Subtract 3 from both sides 3x-3=312-3. Once you subtract get 3x=109.then divide 3 both sides . 3x/3=309/3. To get x= 103.
Now once you have that First number 103. Second number 103+1=104 . And third number 103+2=105. To check add them 103+104+105=312
Answer:
The maximum height of the ball is 256m
Step-by-step explanation:
Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t
At maximum height, the velocity of the ball is zero.
velocity = dh/dt
velocity = -32t + 128
Since v = 0 at maximum height
0 = -32t+128
32t = 128
t = 128/32
t = 4seconds
The maximum height can be gotten by substituting t = 4 into the modelled equation.
h = -16t² + 128t
h = -16(4)²+128(4)
h = -16(16)+512
h = -256+512
h = 256m
12a + 6 + 3a = 15a + 2
15a = 15a - 4
0 = - 4
