Will Mark Brainlest helppp plss

What is 23/100 simplified to

IgorC [24]

The simplified answer is 0.23

4 0

3 years ago

Read 2 more answers

Use the diagram to answer the question

Ivan

Answer:

148

Step-by-step explanation:

a line (cd) is 180°

cba + and must be 180°

180-33=148

8 0

3 years ago

What are the y-intercept of the function f(x)=-2x^2-3x+20?

Serhud [2]

Answer:

$\sf A) \ (-4, \ 0) \ and \ \ [\dfrac{5}{2} , \ 0]$

Explanation:

Given function: f(x) = -2x² - 3x + 20

To find the x-intercepts of a function, f(x) = 0

=================

-2x² - 3x + 20 = f(x)

-2x² - 3x + 20 = 0

-2x² - 8x + 5x + 20 = 0

-2x(x + 4) + 5(x + 4) = 0

(-2x + 5) (x + 4) = 0

-2x + 5 = 0, x + 4 = 0

-2x = -5, x = -4

x = -5/-2,x = -4

x = 5/2, x= -4

Coordinates: (-4, 0), (5/2, 0)

4 0

2 years ago

vic and ben's ages are in the ratio of 3:4. if vic is 12 years old and 3 months old how old is ben in years and months??

posledela

Converting their ages to months:
12*12=144months + 3 months
=147 months
Now, 3 is equivalent to 147 months;
4 will be equivalent to;
4*147/3
= 196 months
= 16 years and 4 months

4 0

3 years ago

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

$a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d$

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

3 years ago

Will Mark Brainlest helppp plss​

Will Mark Brainlest helppp plss