4,6,7,11,15,16 this would be the answer!
Answer:
25 meters
Step-by-step explanation:
Since it is a rectangle, the formula of the perimeter will be 2w+2l.
35(2)+2(x)=120.
2x=50
x=25
The y-intercept is 0 because you can add it to the equation and it won't change your slope.
Part 1) <span>If a cylinder has a surface area of 36 square inches and its radius is 1.6 inches, how tall is the cylinder in inches? π = 3.14. Round to the nearest tenth.
we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height
area of the base==pi*r</span>²*h-----> 3.14*1.6²*h----> 8.0384*h in²
perimeter of the base=2*pi*r---> 2*3.14*1.6----> 10.048 in
surface area=36 in²
so
36=2*[8.0384*h]+10.048*h----> 36=26.1248*h
h=36/26.1248----> h=1.38 in------> h=1.4 in
the answer Part 1) is
h=1.4 in
Part 2) <span>2. A regular hexagonal pyramid has a base area of 29 in^2 and a lateral area of 210 in^2. What is the surface area of the hexagonal pyramid?
surface area of the hexagonal pyramid=2*area of base+lateral area
</span>surface area of the hexagonal pyramid=2*29+210----> 268 in²
the answer part 2) is
surface area=268 in²
Part 3)<span> What is the surface area of the following cylinder in square inches? Use π = 3.14 and round your answer to the nearest tenth.
Radius = 3.2 Height = 4.1
</span>we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height
area of the base==pi*r²*h-----> 3.14*3.2²*4.1----> 131.83 in²
perimeter of the base=2*pi*r---> 2*3.14*3.2----> 20.096in
surface area=2*131.83+20.096*4.1------> 346.05 in²-----> 346.1 in²
the answer Part 3)
surface area=346.1 in²
part 4)<span>A cube has surface area of 150 in^2. How long is each side of the cube?
we know that
surface area of cube=6*b</span>²
where b is the length side of the cube
SA=6*b²------> b=√SA/6------> b=√150/6-----> √25----> b=5 in
the answer Part 4) is
b=5 in
Part 5) <span>A cube has sides that are 3 in. long. What is the surface area of the cube?
</span>we know that
surface area of cube=6*b²
where b is the length side of the cube
b=3 in
so
SA=6*3²----> 54 in²
the answer Part 5) is
SA=54 in²
Part 6) <span>What is the surface area of this box?
Height = 3.5 Length = 6 width = 3
surface area of the box=2*area of the base+perimeter of base*height
area of the base=6*3----> 18 ft</span>²
perimeter of the base=2*[6+3]----> 18 ft
height=3.5 ft
surface area of the box=2*18+18*3.5----> 99 ft²
the answer Part 6) is
SA=99 ft²
Part 7) <span>If a cylinder has a surface area of 59 square inches and its radius is 2.1 inches, how tall is the cylinder in inches? Use π = 3.14. Round to the nearest tenth
</span>
we know that
surface area of the cylinder=2*area of the base+perimeter of the base*height
area of the base==pi*r²*h-----> 3.14*2.1²*h----> 13.8474*h in²
perimeter of the base=2*pi*r---> 2*3.14*2.1----> 13.188 in
surface area=59 in²
so
59=2*[13.8474*h]+13.188*h----> 59=40.8828*h
h=59/40.8828----> h=1.44 in------> h=1.4 in
the answer Part 7) is
h=1.4 in
Part 8) <span>What is the surface area of the cone below, rounded to the nearest tenth? Use π = 3.14.
Height = 12 Radius = 5
surface area of the cone=pi*r</span>²+pi*r*l
r=5 in
h=12 in
l=slant height
l²=r²+h²----> l²=5²+12²----> l²=169-----> l=13 in
surface area of the cone=3.14*5²+3.14*5*13-----> 282.6 in²
the answer Part 8) is
SA=282.6 in²
Part 9) <span>A square pyramid has a surface area of 202 ft^2 and a base area of 53 ft^2. What is the lateral area of the pyramid?
we know that
surface area=area of the base+lateral area
lateral area=surface area-area of the base
lateral area=202-53----> 149 ft</span>²
the answer Part 9) is
LA=149 ft²
Part 10)<span>What is the surface area of the cone below, rounded to the nearest tenth? Use π = 3.14
Radius = 2 Height = 4
</span>
surface area of the cone=pi*r²+pi*r*l
r=2 cm
h=4 cm
l=slant height
l²=r²+h²----> l²=2²+4²----> l²=20-----> l=4.47 cm
surface area of the cone=3.14*2²+3.14*2*4.47-----> 40.63 cm²
the answer Part 10) is
SA=40.6 cm²
Calculation of centripetal acceleration(ac):
we can change our speed in terms of m/s
now, we can plug these values
Calculation of tangential acceleration(at):
now, we can plug values
now, we can simplify it
Calculation of resultant acceleration:
now, we can plug values
.............Answer
