All categories

3 years ago

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

4 is the correct answer.

You might be interested in

What number divided by -7 is equal to -15?

Ilia_Sergeevich [38]

105 is the number
......

4 0

3 years ago

The two spheres above have the same center. One has a radius of 4 cm, and the other has a radius of 5 cm. Approximately how much

frozen [14]

Volume a sphere: [4/3]π(r^3)

Space between the spheres = Volume of the larger sphere - Volume of the smaller sphere

= [4/3]π (R^3) - [4/3π](r^3) = [4/3]π(R^3 - r^3) = [4/3]π {(5cm)^3 - (4cm)^3} =

= 255.5 cm^3

Answer: 255.5 cm^3

8 0

3 years ago

Lena bought 10 pounds of potatoes for $5.50 write the cost as cents per pound

MariettaO [177]

<span>10 pounds of potatoes for $5.50
one pound = $5.50 / 10 = $0.55

answer

it cost 55 cents per pound</span>

5 0

3 years ago

Read 2 more answers

Help math pls its a drag and drop

STALIN [3.7K]

A goes in stock price decreased
B goes in stock price increased
C goes in stock price remained constant
D goes in stock price decreased
E goes is stock price increased

Hope this helps!

7 0

3 years ago

Read 2 more answers

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