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3 years ago

What is a rational number? A.9 B.15 C.28 D.25

2 answers:

6 0

Hi there! :) The answer to your question is all of them. A rational number is a number which can be written as a ratio. Ratios are basically fractions so the numerator and denominator are whole numbers. Whole numbers are 5,49, 688 a number WITH NO fraction. I hope I explained it well and it makes sense. ;)

Hope I helped! =)

Darina [25.2K]3 years ago

4 0

The answer is..........
A. 9

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45 leaf that over 99

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4 years ago

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⎯⎯⎯⎯⎯⎯√square root of 200, 14, 14 over<br> least to greastes

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3 years ago

4/w - 10 = 5<br>this is 8th grade<br>

vovikov84 [41]

Answer:

4/w-10=5

One solution was found :

w = 4/15 = 0.267

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4/w-10-(5)=0

Step by step solution :

Step 1 :

Simplify —

Equation at the end of step 1 :

(— - 10) - 5 = 0

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using w as the denominator :

10 10 • w

10 = —— = ——————

1 w

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

4 - (10 • w) 4 - 10w

———————————— = ———————

w w

Equation at the end of step 2 :

(4 - 10w)

————————— - 5 = 0

Step 3 :

Rewriting the whole as an Equivalent Fraction :

3.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using w as the denominator :

5 5 • w

5 = — = —————

1 w

Step 4 :

Pulling out like terms :

4.1 Pull out like factors :

4 - 10w = -2 • (5w - 2)

Adding fractions that have a common denominator :

4.2 Adding up the two equivalent fractions

-2 • (5w-2) - (5 • w) 4 - 15w

————————————————————— = ———————

w w

Equation at the end of step 4 :

4 - 15w

——————— = 0

Step 5 :

When a fraction equals zero :

5.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

4-15w

————— • w = 0 • w

Now, on the left hand side, the w cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

4-15w = 0

Solving a Single Variable Equation :

5.2 Solve : -15w+4 = 0

Subtract 4 from both sides of the equation :

-15w = -4

Multiply both sides of the equation by (-1) : 15w = 4

Divide both sides of the equation by 15:

w = 4/15 = 0.267

One solution was found :

w = 4/15 = 0.267

Step-by-step explanation:

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3 years ago

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0.44% of 375 please help me

neonofarm [45]

165. Multiply .44 times 375 on the calculator and you’ll get 165

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4 years ago

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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

4 years ago