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4 years ago

The difference of a number x and 5

Mathematics

1 answer:

ser-zykov [4K]4 years ago

3 0

Answer:

x-5

Step-by-step explanation:

"difference" means subtraction. since there isn't a specific number given then we can use x. "and" indicates where the symbols goes.

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The toy shop stocks tricycles and go-carts.

Mandarinka [93]

Total number of wheels in the toy shop = 37 wheels

Number of wheels for tricycles = 3

Number of wheels for go-carts = 5

Since the total number of tricycles and go-carts are not given, we solve this question using Assumption.

Let's assume: We have 4 tricycles and 4 go-carts

(4 x 3) + (4 x 5)

= 12 + 20

= 32 wheels. This assumption is wrong.

Let's assume: We have 4 tricycles and 5 go-carts

(4 x 3) + (5 x 5)

= 12 + 25

= 37 wheels. This assumption is correct.

Therefore, we have 4 tricycles and 5 go-carts in the toy shop.

To learn more, visit the link below:

brainly.com/question/23264573

3 0

3 years ago

Please Help!!!!!!

jonny [76]

Answer:

Step-by-step explanation:

The distance that you travelled in your hummer between Olympia and Spokane is 319.4 miles.

You fill up your gas tank with 28.68 gallons of premium gasoline that costs $2.89 per gallon. This means that the total cost of premium gasoline that you bought would be

2.89 × 28.68 = $82.8852

your gas mileage in miles per gallon is 319.4/28.68 = 11.14 miles per gallon.

Therefore, if it costs $82.8852 to drive 319.4 miles,

then the cost of just one mile in your hummer would be

82.8852/319.4 = $0.26 per mile.

5 0

3 years ago

A group of 4 students need to share 20 paper clips equally ?

harkovskaia [24]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which of the following is equal to the rational expression when x does not equal 3 or 1 x^2-9/(x-1)(x-3)

yuradex [85]

Answer:

b. $\frac{x+3}{x-1}$

Step-by-step explanation:

Given expression is:

$\frac{x^{2}-9 }{(x-1)(x-3)}$

Let us factorize the numerator using the formula:

$a^{2} -b^{2} =(a+b)(a-b)$

$\frac{(x+3)(x-3)}{(x-1)(x-3)}$

Since x ≠ 3, let us cancel the common factor x - 3 in the numerator and in the denominator.

So, $\frac{x^{2}-9 }{(x-1)(x-3)} =\frac{x+3}{x-1}$

7 0

4 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

2 years ago