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Slav-nsk [51]
3 years ago
14

The length of a rectangular box is 2 units less than its width.

Mathematics
1 answer:
AveGali [126]3 years ago
5 0
15 units = length times length minus 2 Length = 5 Width = 2 Perimeter = 16 15 = y * (y -2) Y = 5 5*2 + 3*2 = 16
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Can anybody help me out with this question? (Area)
oksano4ka [1.4K]

Answer: 788.48 feet squared.

Step-by-step explanation:

First, bring the rectangle to its true size. You can do this by multiplying each side by the ratio 8ft/5cm. This is to remove centimetres from the equation.

22 * 8 / 5 = 35.2 feet

14 * 8 / 5 = 22.4 feet

Next, you can multiply the two sides together to get your area, which is 788.48 feet squared.

7 0
4 years ago
Read 2 more answers
X/-4=-1.11 what is x
emmasim [6.3K]

Answer:

4.44 is your answer

Step-by-step explanation:

Isolate the x. Multiply -4 to both sides

(x/-4)(-4) = (-1.11)(-4)

x = (-1.11)(-4)

Multiply

x = 4.44

4.44 is your answer

~

8 0
3 years ago
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Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credi
Oxana [17]

Answer:

Center: (-2, 4)

Radius: 4

Step-by-step explanation:

To find the centre and radius, we require to identify g , f and c

By comparing the coefficients of 'like terms' in the given equation with the general form.

2g = 4 → g = 2 , 2f = -8 → f = -4 and c = 4 → center=(−g,−f)=(−2,4)

radius = √22+(−4)2−4= √4+16−4=4

Center: (-2, 4)

Radius: 4

Hope This Helps! :)

8 0
3 years ago
A light source is located over the center of a circular table of diameter 4 feet. (See picture below) Find the height h of the l
Alex
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum).  If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2))     =0.0962250K   (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.
3 0
3 years ago
A software company is releasing one of their products on a CD. The manufacturer charges a $5,500 setup fee and $1.00 per CD. App
zlopas [31]
Given:
set-up fee 5,500
variable fee 1 per CD.

Cost per CD is $5.50

5.50 - 1= 4.50

5,500 / 4.50 = 1,222 CDs Choice A.

1*1,222 = 1,222
1,222 + 5,500 = 6,722
6,722/1,222 = 5.50
6 0
4 years ago
Read 2 more answers
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