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Gemiola [76]
4 years ago
13

Find the perimeter use 3.14

Mathematics
1 answer:
stich3 [128]4 years ago
8 0
If we can assume those are semi circles, the answer is 5+7.5pi. However, you would need more information if they are not perfect semicircles. 
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Solve the equation for θ, where 0 ≤ θ ≤ 2π.
leonid [27]

Answer:

Theta=1.57rad, 4.71rad, 0rad, 2π rad, π rad and –π rad

8 0
4 years ago
Can someone explain how to solve the system of equation x+y=3 by graphing?
Bess [88]

Answer:

its not possible


Step-by-step explanation:


5 0
3 years ago
Jeremy wants to construct an open box from an 18-inch square piece of aluminum. He plans to cut equal squares, with sides of x i
serious [3.7K]

a.

The volume of the box V = 4x³ - 72x² + 324x

Since the dimensions of the square piece of paper are 18 inches each, and we cut out a length x from each side to give a total length of 2x cut from each side. So, each dimension is L = 18 - 2x.

Since the height of the open box is x and its base is a square, the volume of the open box ix V = L²x

= (18 - 2x)²x

= (324 - 72x + 4x²)x

= 324x - 72x² + 4x³

The volume of the box V = 4x³ - 72x² + 324x

b.

The minimum value of length of the sides of the squares cut from each corner is x = 3 inches.

  • the length of the box, L = 12 inches,
  • the width of the box = 12 inches
  • and the height of the box, x = 3 inches.

Since the volume of the box is 432 cubic inches,

V = 324x - 72x² + 4x³

324x - 72x² + 4x³ = 432

4x³  - 72x² + 324x - 432 = 0

x³  - 18x² + 81x - 108 = 0

A factor of the expression is x - 3

So, x³  - 18x² + 81x - 108 ÷ x - 3 = x² - 15x + 36

So,  x³  - 18x² + 81x - 108 = (x² - 15x + 36)(x - 3) = 0

Factorizing the expression x² - 15x + 36 = 0

x² - 3x - 12x + 36 = 0

x(x - 3) - 12(x - 3) = 0

(x - 3)(x - 12) = 0

So,  x³  - 18x² + 81x - 108 = (x - 3)(x - 3)(x - 12) = 0

So, (x - 3)²(x - 12) = 0

(x - 3)² = 0 and (x - 12) = 0

x - 3 = √0 and x - 12 = 0

x - 3 = 0 and x - 12 = 0

x = 3 twice and x = 12

Since x = 3 is the minimum value, the minimum value of x = 3.

Since the length of the box, L = 18 - 2x

= 18 - 2(3)

= 18 - 6

= 12 inches

The width of the box = L = 12 inches

The height of the box = x = 3 inches.

So,

The minimum value of length of the sides of the squares cut from each corner is x = 3 inches.

  • the length of the box, L = 12 inches,
  • the width of the box = 12 inches
  • and the height of the box, x = 3 inches.

Learn more about volume of a box here:

brainly.com/question/13309609

6 0
2 years ago
A recent study by a major financial investment company was interested in determining whether the annual percentage change in sto
PolarNik [594]

Answer:

c. y? =+4.198 x

Step-by-step explanation:

Hello!

Using the given data you need to estimate the equation of linear regression.

Dependent variable:

Y: Annual percentage change in stock price for a company.

Independent variable:

X: Annual percentage change in profits for the company.

The population regression line equation is:

Y_i= \alpha + \beta X_i + E_i

To estimate the equation you need to find the point estimator for α and β.

The following formulas are the ones to use:

α ⇒ a= y[bar] - bX[bar]

β ⇒ b= (∑xy - [(∑x)(∑y)]/n)/(∑X²-(∑x)²/n)

As you can see you need to make several summatories before calculating the values of a and b:

n= 7

∑x= 36.30

∑x²= 667.09

∑y= 51.60

∑y²= 747.98

∑xy= 560.74

Sample mean of de dependent variable Y[bar]= ∑y/n= (51.60/7)= 7.37

Sample mean of the independent variable X[bar]= ∑x/n= (36.30/7)= 5.19

b= \frac{560.74-\frac{(36.3*51.6)}{7} }{667.09-\frac{(51.60)^2}{7} }

b= 0.6122

a= 7.37-(0.61*5.19)

a= 4.196

The estimated regression equation is:

Y= 4.196 + 0.6122x

I hope it helps!

8 0
3 years ago
A teacher is comparing the test scores of two classes. The range of class 1 is 14, and the interquartile range is 7. The range o
Archy [21]

Class 1 has more variability because both the range and interquartile range are greater than class 2, meaning the data is more spread out.

<h3>What is range and interquartile range?</h3>

The range is the difference between the largest number and the smallest number in a dataset. The interquartile range is the difference between the first quartile and the third quartile of a dataset.

Range and interquartile range are used to measure the spread of a dataset. The higher range and interquartile range is, the greater the spread and variability of the dataset.

To learn more about range, please check: brainly.com/question/26348529

#SPJ1

4 0
3 years ago
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