All categories

3 years ago

HOW CAN ARITHMETIC AND GEOMETRIC SEQUENCES BECOME LINEAR AND EXPONENTIAL FUNCTIONS?

1 answer:

vladimir1956 [14]3 years ago

3 0

A sequence is a set of numbers, called terms, arranged in some particular order. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. A geometric sequence is a sequence with the ratio between two consecutive terms constant.

You might be interested in

5:35-5:52 I need how much minutes from 5:35 from 5:52 please answer

Greeley [361]

Since these both are on the same hour we'll ignore the 5. We end up having :35 and :52. We'll subtract 35 from 52 which gives us 17. The answer to this is 17 minutes.

8 0

3 years ago

Read 2 more answers

30% of 60 is <br><br> 30%of ____ is 60

ololo11 [35]

You know that 3 times 2 is 6 so 30'times 20 equals 60

6 0

3 years ago

Read 2 more answers

HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

MA_775_DIABLO [31]

Answer:

m<3 = 97

m<10 = 97

m<9 = 83

m<5 = 97

m<7 = 83

m<16 = 97

Step-by-step explanation:

m<3 = m<2 because of vertical angles

m<10 = m<2 because of corresponding angles

m<9 + 97 = 180

m<9 = 83

m<5 + 83 = 180

m<5 = 97

m<7 = m<6 because of vertical angles

m<16 = m<5 because of alternate exterior angles.

4 0

3 years ago

The graph of which function has a minimum located at (4, –3)?

valentinak56 [21]

Answer:

None of the options is the answer to the question

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

case A) we have

$f(x)=x^{2}+4x-11$

In this case the x-coordinate of the vertex will be negative

therefore

case A is not the solution

case B) we have

$f(x)=-2x^{2}+16x-35$

This case is a vertical parabola open downward (the vertex is a maximum)

The vertex is the point $(4,-3)$ <u>but is not a minimum</u>

see the attached figure

therefore

case B is not the solution

case C) we have

$f(x)=x^{2}-4x+5$

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-5=x^{2}-4x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-5+4=(x^{2}-4x+4)$

$f(x)-1=(x^{2}-4x+4)$

Rewrite as perfect squares

$f(x)-1=(x-2)^{2}$

$f(x)=(x-2)^{2}+1$ --------> vertex form

The vertex is the point $(2,1)$

therefore

case C is not the solution

case D) we have

$f(x)=2x^{2}-16x+35$

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-35=2x^{2}-16x$

Factor the leading coefficient

$f(x)-35=2(x^{2}-8x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-35+32=2(x^{2}-8x+16)$

$f(x)-3=2(x^{2}-8x+16)$

Rewrite as perfect squares

$f(x)-3=2(x-4)^{2}$

$f(x)=2(x-4)^{2}+3$ --------> vertex form

The vertex is the point $(4,3)$

therefore

case D is not the solution

The answer to the question will be the function

$f(x)=2x^{2}-16x+29$

7 0

4 years ago

Read 2 more answers

Craig practiced his drum 4 hours more than nicki . Nicki practiced his drums 5 hours less than marshal marshal practiced his dru

abruzzese [7]

Craig practiced 15 hours

8 0

3 years ago