Combine any like terms in the expression. If there are no like terms, rewrite the expression.

a bicycle tire is 28 inches in diameter. approximately how far does the bicycle move forward each time the wheel goes around? (u

belka [17]

ANSWER

$88 \: inches$

EXPLANATION

The given bicycle has a tyre that is 28 inches in diameter.

How far the bicycle moves forward each time the wheel goes around is the circumference of the bicycle tyre.

This is calculated using the formula:

$C =\pi \: d$

We substitute the diameter and

$\pi = \frac{22}{7}$

$C = \frac{22}{7} \times 28$

This simplifies to

$C =22 \times 4$

$88 \: inches$

8 0

3 years ago

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Material for a dress costs £5.50 per metre. find the cost of 10cm and 70cm

Zielflug [23.3K]

Answer:

The cost of 10 cm and 70 cm are £0.55 and £3.85 respectively.

Step-by-step explanation:

Material for a dress costs £5.50 per meter.

We know that, 1 meter = 100 cm.

Suppose, the cost of 10 cm material is $x$ and the cost of 70 cm material is $y$

Now, according to the ratio of "cost" to the "length of material" , we will get.......

$\frac{x}{10} =\frac{5.50}{100}\\ \\ 100x=10*5.50=55\\ \\ x=\frac{55}{100}=0.55$

and

$\frac{y}{70} =\frac{5.50}{100}\\ \\ 100y=70*5.50=385\\ \\ x=\frac{385}{100}=3.85$

So, the cost of 10 cm material is £0.55 and the cost of 70 cm material is £3.85

5 0

3 years ago

Which equation represents a proportional relationship?

iren2701 [21]

The equation that represents a proportional relationship is:

y = 32x

Because:

y/x = 32

So that implies that y is directly proportional to x.

Note that other equations represent a proportional relationship in some form, only that the relationship y = 32x is the only direct proportional relationship.

5 0

3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago

Broccoli cost $3.14 a pound at the market.

pav-90 [236]

7.85

explanation

you multiply the two numbers together

4 0

2 years ago

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