Find the angle measures in the regular polygon.

the perimeter of a square is 20 ft if you increase the length of the square by 2 feet and decrease the width by 1ft what is the

IrinaVladis [17]

Answer:

28 ft squared

Step-by-step explanation:

The 4 sides of a square are equal in length, so if the perimeter is 20ft that means each side would have to be 5ft, (20/4 = 5). If you increase the length by 2ft, one of the sides would now become 7ft (5+2 = 7). If you decrease the length of the other side by 1ft, the length would now be 4ft (5-1=4). Now to find the area, all you have to do is length * width. So you would do 7ft*4ft which is 28ft squared.

4 0

3 years ago

Adjacent angles and linear pairs can you help?

Natalija [7]

Answer:

Step-by-step explanation:

x + 13 + 4x + 2 = 180

Add the like terms,

5x + 15 = 180

Subtract 15 form both sides

5x + 15 - 15 = 180 -15

5x = 165

Divide both sides by 5

5x/5 = 16/5

x = 33

8 0

3 years ago

The length of a rectangular playing field is 8 yards longer than quadruple the width. If the perimeter of the rectangular playin

kipiarov [429]

Answer:

Step-by-step explanation:

L=4W+8

P=2(L+W)

P=2(5W+8) and P=476

2(5W+8)=476

5W+8=238

5W=230

W=46yds

L=4(46)+8=192yds

So the dimensions are 46X192 yards

5 0

3 years ago

According to industry sources, online banking is expected to take off in the near future. The projected number of households (in

Airida [17]

Answer:

(a) The least-square regression line is: $y=4.662+2.709x$ .

(b) The number of households using online banking at the beginning of 2007 is 31.8.

Step-by-step explanation:

The general form of a least square regression line is:

$y=\alpha +\beta x$

Here,

y = dependent variable

x = independent variable

α = intercept

β = slope

(a)

The formula to compute intercept and slope is:

$\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \end{aligned}$

The values of ∑X, ∑Y, ∑XY and ∑X² are computed in the table below.

Compute the value of intercept and slope as follows:

$\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 68.6 \cdot 55 - 15 \cdot 218.9}{ 6 \cdot 55 - 15^2} \approx 4.662 \\ \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 6 \cdot 218.9 - 15 \cdot 68.6 }{ 6 \cdot 55 - \left( 15 \right)^2} \approx 2.709\end{aligned}$