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1 year ago

the diagonal length of a square window is 40 cm. find the area of the window. Btw any help is appreciated ! :)

1 answer:

7 0

Ok so area is length times width. A square has the same sides all around, which is called a quadrilateral. So you would plug in 40 for length and width. 40x40= 1,600 cm^2 <———— is your answer.

Hope this helped :)

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A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below.

sergiy2304 [10]

The equation of a parabola with vertex at (h,k) is
y=a(x-h)²+k

vertex isi at (0,0)

y=a(x-0)²+0
y=a(x)²
y=ax²
find a

we see that one point is (14,-74)
x=14 and y=-74

-74=a(14²)
-74=196a
divide both sides by 196
-37/98=a

the equation is

$y= \frac{-37}{98} x^2$

6 0

2 years ago

Read 2 more answers

Instructions:Select the correct answer from each drop-down menu. The sides of ∆ABC are 30 units, 40 units, and 60 units long. Th

andriy [413]

Answer:

The sides of ∆ABC are 30 units, 40 units, and 60 units long.

The perimeter will be = $30+40+60=130$ units

Given is that ∆ABC and ∆XYZ are similar and the corresponding sides of ∆XYZ are 'r' times as long as the sides of ∆ABC.

So, perimeter of ∆XYZ will be : $130\times r=130r$ units.

----------------------------------------------------------------------------------------

The area of ∆ABC is n square units, so area of ∆XYZ will be :

$\frac{1}{2}(r\times height)(r\times base)$

= $n\times r\times r$

= $nr^{2}$ square units.

8 0

2 years ago

A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained

sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

Sample 1

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.

Sample 2

Min= 50.32

Max= 52.56

Mean= 51.16

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both the max and min values of the sample are within the working range without warning.

Sample 3

Min= 50.25

Max= 53.12

Mean= 51.83

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.

Sample 4

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

Both min and max values surpass the warning interval but do not reach action levels.

Sample 5

Min= 50.35

Max= 52.71

Mean= 51.37

Without warning 50.22<X[bar]<52.78