Ask question

Ask question

All categories

2 years ago

5

The nearest star to the earth is

0%7Bx%20%7D%5E%7B2%7D%20" id="TexFormula1" title=" {2 \times + 4}^{2} = {x }^{2} " alt=" {2 \times + 4}^{2} = {x }^{2} " align="absmiddle" class="latex-formula">

1 answer:

dalvyx [7]2 years ago

7 0

The nearest start of the earth is Alpha Centauri and the answers to the math question is X= 4 Square root 2

You might be interested in

Helpppppppppppppppppp

topjm [15]

Answer:

B

Step-by-step explanation:

The question has no division answer. That makes C and D incorrect.

A for some reason is backwards. It makes no sense to use that. You are left with B. The reason A or B is correct is that the burn of calories must get larger with an increase in hours.

7 0

3 years ago

Find the y-intercept of a line that passes through (-2,6) and has a slope of -5

a_sh-v [17]

Answer:

its gonna be -4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the arc length of the bolder arc<br> A) 283.5 mi<br> C) 733.0 mi<br> B) 10.2 mi<br> D) 73.3 mi

statuscvo [17]

Answer:

d its d :)

Step-by-step explanation:

i already did that

7 0

3 years ago

Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal

pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and $\frac{11}{6}$ .

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form $f(x) = 0$ based on the following formula:

$x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}$

Where:

$x_{i}$ - i-th Approximation, dimensionless.

$x_{i+1}$ - (i+1)-th Approximation, dimensionless.

$f(x_{i})$ - Function evaluated at i-th Approximation, dimensionless.

$f'(x_{i})$ - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be $f(x) = x^{2}-8$ and $f'(x) = 2\cdot x$ , the resultant expression is:

$x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}$

First iteration: ( $x_{1} = 2$ )

$x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}$

$x_{2} = 2 + \frac{4}{4}$

$x_{2} = 3$

Second iteration: ( $x_{2} = 3$ )

$x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}$

$x_{3} = 2 - \frac{1}{6}$

$x_{3} = \frac{11}{6}$

7 0

3 years ago

there is an abondon house with a fish tank in it that has 7 fish 4 of the 7 die and 2 fish get eaten by another fish haw many fi

soldier1979 [14.2K]

There is one fishie left.
have a nice night
bye

5 0

3 years ago