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3 years ago

7

A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18 \pi18π18, pi

meters.
What is the area AAA of the cross section of the column?
Give your answer in terms of pi.

1 answer:

katrin [286]3 years ago

5 0

The area of the cross section of the column is $18 \pi \ m^2$

Explanation:

Given that a building engineer analyzes a concrete column with a circular cross section.

Also, given that the circumference of the column is $18 \pi$ meters.

We need to determine the area of the cross section of the column.

The area of the cross section of the column can be determined using the formula,

$Area= \pi r^2$

First, we shall determine the value of the radius r.

Since, given that circumference is $18 \pi$ meters.

We have,

$2 \pi r=18 \pi$

$r=9$

Thus, the radius is $r=9$

Now, substituting the value $r=9$ in the formula $Area= \pi r^2$ , we get,

$Area = \pi (9)^2$

$Area = 81 \pi$

Thus, the area of the cross section of the column is $18 \pi \ m^2$

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Answer:

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3 years ago

Solve (x+y) if x=-1/2 and y=-3

bixtya [17]

Answer:

$-3\frac{1}{2}$

Step-by-step explanation:

Solve

$(x+y)$

For x=-1/2 and y=-3

substitute the value of x and the value of y in the expression above

$x+y=(-\frac{1}{2})+(-3)$

Remember that

$(+1)(-1)=-1$

so

$x+y=-\frac{1}{2}-3=\frac{-1-6}{2}=-\frac{7}{2}$

Convert to mixed number

$-\frac{7}{2}=-\frac{6}{2}-\frac{1}{2}=-3\frac{1}{2}$

7 0

3 years ago

Suppose you are bending stainless steel wire to make the earring shown below. The

lakkis [162]

Given parameters:

Length of earring hook = 1.5cm

Length of given wire = 45cm

Dimension of the triangle = 3.2cm, 3.2cm and 2.1cm

Unknown:

Number of earrings that can be made from the wire = ?

To solve this problem, we need to calculate the total length of the earring we want to make.

Total earring length = Perimeter of earring + length of earring hook

= 3.2 + 3.2 + 2.1 + 1.5

= 10cm

Now the length of the wire is 45cm,

Number of earrings = $\frac{45}{10}$ = 4.5 earrings

But this is not reasonable, therefore, we can only forge 4 complete earrings from the given length with 5cm of wire remaining.

The number of earrings that can be made is 4.

4 0

3 years ago

In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co

sukhopar [10]

Answer:

a. proportions have not changed significantly

Step-by-step explanation:

Given

Business College= 35 %

Arts College= 35 %

Education College = 30%

Calculated

Business College = 90/300= 9/30= 0.3 or 30%

Arts College= 120/300= 12/30= 2/5= 0.4 or 40%

Education College= 90/300= 9/30 = 0.3 or 30%

First we find the mean and variance of the three colleges using the formulas :

Mean = np

Standard Deviation= s= $\sqrt{npq\\}$

Business College

Mean = np =300*0.3= 90

Standard Deviation= s= $\sqrt{npq\\}$ = $\sqrt{0.3*0.7*300}$ = 7.94

Arts College

Mean = np =300*0.4= 120

Standard Deviation= s= $\sqrt{npq\\}$ = $\sqrt{0.4*0.6*300}$ = 8.49

Education College

Mean = np =300*0.3= 90

Standard Deviation= s= $\sqrt{npq\\}$ = $\sqrt{0.3*0.7*300}$ = 7.94

Now calculating the previous means with the same number of students

Business College

Mean = np =300*0.35= 105

Arts College

Mean = np =300*0.35= 105

Education College:

Mean = np =300*0.3= 90

Now formulate the null and alternative hypothesis

Business College

90≤ Mean≥105

Arts College

105 ≤ Mean≥ 120

Education College

U0 : mean= 90 U1: mean ≠ 90

From these we conclude that the proportions have not changed significantly meaning that it falls outside the critical region.

8 0

3 years ago

Which is the last sentence of the proof? because f + e = 1, a2 + b2 = c2. Because f + e = c, a2 + b2 = c2. Because a2 + b2 = c2,

lyudmila [28]

The latest sentence of proof by using the triangle congruence method we can conclude that f+e=c, $a^{2} + b^{2} = c^{2}$ .

Given that ΔABC and ΔCBD are right triangles

Therefore, one angle of both triangle is of 90°

So, ∠B is same in both triangles. Hence, by suing the Angle-Angle theorem rule we can conclude that both ΔABC and ΔCBD are similar.

Consequently, ∠A is same in both triangles. Hence, by suing the Angle-Angle theorem rule we can conclude that both ΔABC and ΔCBD are similar.

Similarly, When two triangles are similar then their corresponding angles are equal and their corresponding sides are also equal.

Therefore , the two proportions can be rewritten as

a² = cf ( equation 1 )

b² = ce ( equation 2 )

By adding b² on both side of equation 1 then we can write equation 1 as

$a^{2} + b^{2} = b^{2}+cf$

$a^{2} + b^{2} = ce+cf$

Because b² and ce are equal and substitute on the right side of equation 1

Using the converse of distributive property in above equation then we get a new equation. That is,

$a^{2} +b^{2} = c(f+e)$

Because distributive property is a(b+c)=a(b+ac)

a² + b² = c²

Because e + f = c²

To learn more about triangle congruency: brainly.com/question/1675117

#SPJ4

6 0

6 months ago