All categories

3 years ago

5.1million x 2.9 thousand

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

4 0

Answer:

fourteen billion seven hundred ninety millio

You might be interested in

A hiker cycles 9 miles in 2/3 of an hour. Make a chart and fill it in to show how far he will bike if he maintains the same pace

Arada [10]

Answer:

2 hours: 27 miles, 3 hours: 40.5 miles, 3.5 hours: 47.25 miles

Step-by-step explanation:

1.) If a hiker hikes 9 miles in 2/3 of an hour, he will hike 4.5 (Half of 2/3 is 1/3 and half of 9 is 4.5) miles in 1/3 of an hour and 13.5 miles per hour. (1/3 times 3 gives us a whole and 4.5 times 3 is 13.5)

2.) 2 hours: 13.5 * 2 = 27 miles

3.) 3 hours: 13.5 * 3 = 40.5 miles

4.) 3.5 hours: 13.5 * 3.5 = 47.25 miles

6 0

3 years ago

Read 2 more answers

Round 1199.28856995 to the nearest ten,

Wittaler [7]

1199.3 (I'm assuming you mean round to the nearest tenth)

5 0

3 years ago

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is cl

gladu [14]

Consider the attached ellipse. Let the sun be at the right focus. Then perihelion is at right vertex on the x-axis and aphelion is at the left vertex on the x-axis.

The distances:

from perihelion to the sun in terms of ellipse is a-c;
from aphelion to the sun in terms of ellipse is a+c.

Then

$\left\{\begin{array}{l}a-c=741,000,000\\a+c=817,000,000\end{array}\right.$

Add these two equations:

$2a=1,558,000,000 \\ \\a=779,000,000$

and subtract first equation from the second:

$2c=76,000,000 \\ \\c=38,000,000.$

Note that $b=\sqrt{a^2-c^2},$ thus

$b=\sqrt{779,000,000^2-38,000,000^2}=\sqrt{605,397,000,000,000,000}.$

The equation for the planet's orbit is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Rightarrow \dfrac{x^2}{606,841,000,000,000,000}+\dfrac{y^2}{605,397,000,000,000,000}=1.$

7 0

3 years ago

Which expression has both 8 and n as factors?<br><br> 8 / n 8 + n 8n 8 - n

Sidana [21]

Answer:

Step-by-step explanation:

3 0

2 years ago

Prove that for all n in the naturals, n(n+1)(n+2) is divisible by 3

DerKrebs [107]

By induction:

It's true for $n=1$ , since $1\cdot2\cdot3$ clearly contains a factor of 3.

Suppose it's true for $n=k$ , that $k(k+1)(k+2)$ is divisible by 3. Then

$(k+1)(k+2)(k+3)=\dfrac{k(k+1)(k+2)(k+3)}k=\dfrac{3m(k+3)}k$

where $m$ is an integer. This reduces to

$\dfrac{3m(k+3)}k=3m+9\dfrac mk$

and both terms are clearly multiples of 3. We know that $\dfrac mk$ is an integer since we had set $m=k(k+1)(k+2)$ previously, which implies $m$ is a multiple of $k$ . So the statement is true.

6 0

3 years ago