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3 years ago

What’s the numerator for the following rational expression s/t+7/t=?/t

Mathematics

1 answer:

avanturin [10]3 years ago

4 0

Answer:

s+7 is numerator of the given expression.

Step-by-step explanation:

We have bee given a rational expression:

$\frac{p}{q}$ is the general representation of the Rational expression where, p is numerator and q is denominator.

So, our given expression is: $\frac{s}{t}+\frac{7}{t}$

We will take the LCM so, given expression can be rewritten as:

$\frac{s+7}{t}$ (1)

Comparing (1) with general rational expression written above

p=s+7 which is numerator of the given expression.

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If Q is the midpoint of PR, find x when PR=2x and PQ=2x-12.

frozen [14]

Answer: x=12

Step-by-step explanation:

You start with: $2x\neq 2x-12$

The segment PQ is half of PR, so to make both sides equal the PQ side must be doubled.

$2x=4x-24$

Then all x-values should be put on one side of the equation.

$2x-4x=-24$

Combine like terms.

$-2x=-24$

Then simplify. $x=12$

6 0

3 years ago

How do you solve this to get to the 2^12?

Ivan

2x2x2x2x2x2x2x2x2x2x2x2

2 multiplied 12 times.

8 0

4 years ago

A one-dollar bill is 1 × 10–4 meter thick. How tall is a stack of 500 one-dollar bills?

Anna [14]

Answer:

4996

Step-by-step explanation:

Help this help:)

500x10-4= 4996

6 0

3 years ago

Read 2 more answers

You have r rare coins, consisting of p pennies and n nickels.

Marysya12 [62]

9514 1404 393

Answer:

a. 38

b. 95

c. 57

Step-by-step explanation:

In this context, you can consider "of" to mean "times." It can also be helpful to think of % as meaning /100.

a. p = 20% × 190

p = 20/100 × 190 = 38 . . . . you have 38 pennies

b. 190 = 200% × r

190 = 200/100 × r . . . . replace % with /100

190 = 2 × r . . . . . . . . . simplify the fraction

190/2 = r = 95 . . . . you have 95 rare coins

c. n = 60% × r

n = 60/100 × 95 = 57 . . . . you have 57 nickels

8 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

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

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

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago