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kap26 [50]
3 years ago
14

Is 14 squared bigger than 3.14

Mathematics
1 answer:
zloy xaker [14]3 years ago
6 0

Answer:

Yes, 14 squared is definitely bigger than 3.14. When you square a number, you multiply it by itself, and even if you didn't do that, 14 would still be a bigger number than 3.14.

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What is the equation of the line that passes through the points (-9,-6) and (-9, -8)?​
andrey2020 [161]

Answer:

x=-9

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-8-(-6))/(-9-(-9))

m=(-8+6)/(-9+9)

m=-2/0

undefined

x=-9

7 0
3 years ago
7%of what length is 200 feet?
Aneli [31]

Step-by-step explanation:

Let the length be x

7% of x = 200ft

7/100 × x = 200

7x/100 = 200

7x = 200 × 100

7x = 20000

x = 20000/7

x = 2,857.14

8 0
3 years ago
Read 2 more answers
What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

                                           = 5.2 feet

r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

                            = 3.14 x (2.6)^{2}

                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

                                           = 21.2264 - 19.625

                                           = 1.6014

The cross sectional area of the cement pipe is 1.60 ft^{2}.

6 0
3 years ago
Christopher Columbus set sail from Spain in 1492. In what year did the Civil War end?
horsena [70]

Answer:

idk when did it end huhuhuhuhuhuhuhuhu

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Two forces with magnitudes of 90 and 50 pounds act on an object at angles of 30° and 160°, respectively. Find the direction and
Stels [109]
Force 1, F1 = 90, angle 30°

Force 2, F2 = 50, angle 160°

F1 = 90 cos(30) i + 90 sin (30) j
F2 = 50 cos (160) i + 50 sin (160) j

F1 = 90*0.866 i + 90*0.5 j
F2 = 50*(- 0.940) i + 50*0.342 j

F1 = 77.94 i + 45 j
F2 = -47 i +17.10 j

Resultant force, Fr = F1 + F2

Fr = [77.94 - 47.00] i +[45 +17.10]j = 30.94 i + 62.10 j

Magnitude = √[30.94 ^2 + 62.10^2] = 69.38 pounds

Direction  = arctan[62.10/30.94] = 63.52 °


7 0
3 years ago
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