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Ivanshal [37]
2 years ago
6

Enter the value that makes the given equation true. Y+396.23 =574.54

Mathematics
1 answer:
Levart [38]2 years ago
8 0
574.54-396.23= 178.31
So Y=178.31
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Please help !!
olga2289 [7]
If you have learned how to find the line of best fit manually, then you can do it that way. Perhaps you may want to just find a line that can connect at least two of the points and I believe that that line will be able to represent the other points because, in general, the points are pretty close to one another. 

If you don't want to do it manually and have a graphing calculator (which I recommend) then you can use that to find the line of best fit (and if you want then you can see how precise your points are with your r^2 value). Or there is a website (http://illuminations.nctm.org/Activity.aspx?id=4186), which you can use to help you to find the equation of that particular line. 

Once you have that done, then you can substitute 2009 for the x value in the equation and then see what y value the equation produces. That will then be your answer :) 
4 0
3 years ago
D is the midpoint of CE. C has coordinates
tiny-mole [99]

Answer:

you need to be more specific like which ones are x or y

Step-by-step explanation:

ask it again or comment the specifics

4 0
3 years ago
A large pizza at angelo's pizzeria has a diameter of 14 inches. what is the area of the pizza? use 3.14 for π.round to the neare
Korolek [52]
The answer is 153.9 inches squared
7 0
2 years ago
Given the sequence 2, 5, 8, 11, 14, ..., which term is 59? (Hint: find n.)
____ [38]
N = 3  ( the next number is 3 plus the previous number
 sequence formula = an = 3(n-1) +2

so for term the term that =59

59 = 3(n-1) +2

distribute:
59 = 3n -3+2

59 = 3n -1

60 = 3n
n = 60/3
n = 20

59 is the 20th term



8 0
3 years ago
The value of a car depreciates by 35% per year. Work out the current value of a car bought 2 years ago for £20000
Rzqust [24]

First year: the depreciation is (35/100) x 20000 = £7000; now the value of the car is £20000 - £7000 = £13000;

Second year: the depreciation is (35/100) x 13000 = £4550; the current value of the car is £13000 - £4550 = £8450.

6 0
3 years ago
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