All categories

3 years ago

Find the first 5 terms of the sequence: a1 = 500, an = (an-1)/5.

Mathematics

1 answer:

Svet_ta [14]3 years ago

8 0

Since formula is already give, use it to calculate
500, 100, 20, 5, 1

You might be interested in

Enter your answer in the box.

Marizza181 [45]

the area is about 41.033ft2

7 0

4 years ago

The following proof shows an equivalent system of equations created from another system of equations. Fill in the missing reason

aev [14]

Answer:

Equation 2 of system divided by 2 on both sides then we get second equation of system 2.

Step-by-step explanation:

We are given two equations

$2x+2y=14$

$-4x+6y=12$

We are given a proof which shows an equivalent system of equations created from another system of equations

We have to fill the correct reason in the blank space

Statement: $2x=2y=14$

$-4x+6y=12$

Reason: Given

Statement : $2x=2y=14$

$-2x+3y=6$

Reason : equation 2 of system I divided by 2 on both sides then we get second equation of system 2.

When we divide by 2 then the solution does not change and x and y - intercepts are not change.So, the two systems are equivalent.

3 0

3 years ago

Solve for X . Please help !! :)

Orlov [11]

Answer:

Step-by-step explanation:

thats the answer

x=5

2x=10

x√3=√170

6 0

3 years ago

- Polynomial Functions -For each function, state the vertex; whether the vertex is a maximum or minimum point; the equation of t

vladimir2022 [97]

EXPLANATION

Given the function f(x) = (x-6)^2 + 1

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$=x^2-12x+37$ $\mathrm{The\: parabola\: params\: are\colon}$ $a=1,\: b=-12,\: c=37$ $x_v=-\frac{b}{2a}$ $x_v=-\frac{\left(-12\right)}{2\cdot\:1}$ $\mathrm{Simplify}$ $x_v=6$ $y_v=6^2-12\cdot\: 6+37$

Simplify:

$y_v=1$

$\mathrm{Therefore\: the\: parabola\: vertex\: is}$ $\mleft(6,\: 1\mright)$ $\mathrm{If}\: a$ $\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}$ $a=1$ $\mathrm{Minimum}\mleft(6,\: 1\mright)$ $\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$y=x^2-12x+37$ $\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}$ $a=1,\: b=-12$ $x=\frac{-\left(-12\right)}{2\cdot\:1}$ $\mathrm{Refine}$

Axis of simmetry : x=6

The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.

3 0

1 year ago

1.What is the Line of Best Fit for this data? Perform a linear Regression on this table to

Ber [7]

Answer:

Step-by-step explanation:

5 0

3 years ago