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igomit [66]
3 years ago
14

How does the function g(x)= - (x - 1) compare to the parent function f(x) = x ?

Mathematics
1 answer:
grandymaker [24]3 years ago
3 0
Reflection across x axis and move 1 unit to the right
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Write an equation for the line that passes through (2, -2) and (4, -1)​
Dahasolnce [82]

Answer:  y=1/2x-3

Step-by-step explanation:

Slope-intercept form:y=mx+b

m=y2-y1/x2-x1

-1-(-2)/4-2=1/2

y=1/2x+b

-1=1/2(4)+b

-1=2+b

b=-3

6 0
3 years ago
A ribbon is 4 m long. How many pieces each 30 cm long can be cut from the ribbon?​
EleoNora [17]

Answer:

13 pieces

Step-by-step explanation:

First we convert 4m into cm :

1m = 100cm

4m = 400cm

Now we need to figure out how many 30s go into 400 :

400 ÷ 30 = 13 remainder 10

So 13 pieces can be cut from the ribbon

Hope this helped and have a good day  

5 0
2 years ago
Read 2 more answers
8
tia_tia [17]

Answer:

8-6i

Step-by-step explanation:

Multiply out (3-i)^2 in order to find out which expression it is equivalent to.

1) Since the whole quantity is squared, write it out as (3-i)(3-i).

(3-i)^2\\(3-i)(3-i)

2) Multiply binomials by using the FOIL method. Multiply the terms that are listed first in each binomial, then the ones that are listed outermost when looking at both binomials, then innermost, and finally the last terms listed in each binomial. Simplify and combine like terms.

(3-i)(3-i)\\9 - 3i - 3i + i^2

9 - 6i + i^2

3)  i = \sqrt{-1}, so i ^2 must be (\sqrt{-1} )(\sqrt{-1}). Thus, i^2 is equivalent to -1. Knowing this, simplify and combine like terms.

9 - 6i + ((-\sqrt{1} )(\sqrt{-1} ))

9 - 6i -1\\8 - 6i

Thus, it is equivalent to  8 - 6i .

6 0
3 years ago
The vertices of triangle JKL, shown below, are J(2, 1), K(4, 4), and L(6, 2). If triangle JKL is translated 2 units to the left
julia-pushkina [17]

Answer:

L=(4,-2)

Step-by-step explanation:Left = -2 units and down = -4 units, left is x direction and down is y direction. (x,y).

8 0
2 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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