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aleksklad [387]
4 years ago
13

The distance from the normal hairline to the base of the nose is equal to to the distance from the base of the chin to the

Mathematics
1 answer:
inn [45]4 years ago
8 0

The answer is eyebrow. According to a research, the mean distance from chin to nose was 6.96 cm. The mean distance from nose to eyebrows was 6.55 cm. The mean space from eyebrows to hairline was 6.8 cm. while, the space from chin to hairline was 20.3 cm.

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Question in the picture
Irina18 [472]

Answer:

The mistake is that 16 was added to the other side of the equal sign instead of subtracted in step 3. Then b^2 should equal 48 and the rest is right.

Step-by-step explanation:

7 0
4 years ago
A rectangular prism has a volume of 320 cubic cm,and a height of 8 cm . What is the area of its base
gregori [183]

Volume =length *breadth *height

Volume =area of the cross section *height

320=area of the cross section * 8

Area of the cross section =320/8

=40cm2

4 0
3 years ago
If the mean of this data is 95, find the missing data 103,99,90,86,105,96 what is the missing data
d1i1m1o1n [39]
Let, the missing number = x
Then, 103 + 99 + 90 + 86 + 105 + 96 + x / 7 = 95
579+x = 665
x = 665 - 579
x = 86

In short, Your Answer would be 86

Hope this helps!
3 0
3 years ago
PLEASE HELP ME. ANSWER THIS FAST 20 POINTS FOR IT!
9966 [12]

Answer:

(<) this means less than 

(>) this means greater than

So since the question doesn't say anything about -4°C we can automatically cross out A and B 

They are both negative! but when you have a negative. Its going to be less than a positive 

So this means 4°C is greater than -5°C 

or in other words

D.) 4°C > -5°C

hope this helps! 

and don't forget 2 

MARK ME BRAINIEST! 

=)

Step-by-step explanation:

5 0
3 years ago
You have 800 feet of fencing and you want to make two fenced in enclosures by splitting one enclosure in half. What are the larg
katovenus [111]
Problema Solution

You have 800 feet of fencing and you want to make two fenced in enclosures by splitting one enclosure in half. What are the largest dimensions of this enclosure that you could build?

Answer provided by our tutors

Make a drawing and denote:


x = half of the length of the enclosure


2x = the length of the enclosure


y = the width of the enclosure


P = 800 ft the perimeter


The perimeter of the two enclosures can be expressed P = 4x + 2y thus


4x + 3y = 800


Solving for y:

........

click here to see all the equation solution steps

........

y = 800/3 - 4x/3


The area of the two enclosure is A = 2xy.


Substituting y = 800/3 - 4x/3 in A = 2xy we get


A = 2x(800/3 - 4x/3)


A =1600x/3 - 8x^2/3


We need to find the x for which the parabolic function A = (- 8/3)x^2 + (1600/3)x has maximum: 


x max = -b/2a, a = (-8/3), b = 1600/3


x max = (-1600/3)/(2*(-8/3))


x max = 100 ft


y = 800/3 - 4*100/3


y = 133.33 ft


2x = 2*100


2x = 200 ft

3 0
3 years ago
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