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4 years ago

What is the equation for the blue line?

Mathematics

2 answers:

grandymaker [24]4 years ago

5 0

Answer:

f(x) = -1/2x + 2

Step-by-step explanation:

f(x) = ax + b

x = 0 , y = 2

y = 0 , x = 4

$a = \frac{y'' - y'}{x'' - x'}\\a= \frac{2 - 0}{0 - 4} \\a= \frac{2}{-4} = \frac{-1}{2}$

$0 = \frac{-1}{2} . 4 + b\\0 = -2 + b\\b = 2$

f(x) = -1/2x + 2

vladimir1956 [14]4 years ago

4 0

I personally like to start by finding the y intercept first. When x is 0 y is 2

Then to find the actual slope, firstly the line is going in a negative way so the slope is negative. You then do rise over run and you get -1/2

The slope is y=-1/2 +2

Hope this helps!!!!

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I need help quickly!!!!

Mademuasel [1]

A) Yes, it starts with the same coordinates. because it start as (0,0)
B) No, it because it (6,3)
C) No, because the line segment goes higher and higher.
D) No, because the domain and range dont repeat.
E) I dont exactly know sorry :/

8 0

3 years ago

Dr. Graham was curious as to whether memory is affected by position during study. Participants were randomly assigned to study a

lina2011 [118]

Answer:

Dependent variable: Memory

Independent variable: Position

Step-by-step explanation:

We are given the following in the question:

Dr. Graham was curious as to whether memory is affected by position during study.

He wants to check the effect of position(either laying down on a bed or sitting up at a desk) effects the memory(list of 50 words).

Dependent and independent variable:

Dependent variable is affected by the changes in the independent variable.
Dependent variable are also known as response variable.
Independent variables are the free variables.

For the given study

Dependent variable:

Memory that is determined by the recalling power of list of 50 words.

Independent variable:

Position that is either laying down on a bed or sitting up at a desk

3 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1