If x+y=3 , x²+y²=5 then xy is what

how many more miles is the distance from mars to the sun than the distance from tge earth to the sun ?

enot [183]

Answer:

Mars orbits the Sun at an average distance (semi-major axis) of 228 million km (141.67 million mi), or 1.524 astronomical units (over one and a half times the distance between Earth and the Sun).

Step-by-step explanation:

7 0

3 years ago

Please help I'm stuck on this questions thanks

polet [3.4K]

So its asking for basically the percentage of the first number out of the second.

3. 25/50 = 50%
4. 125/75 = 167%
5. 32/28 = 114%
6. 7/10 = 70%

Hope this helped! :)

6 0

3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago

There were 27,000 people at a ball game in Los Angeles. The day's receipts were $204,000. How many people paid $13 for reserved

Studentka2010 [4]

21000 people paid for general admission and 6000 paid for reserved seats. This is solved by making 2 equations. Out of 27000 people who were at game, x of them paid for general admission and y for reserved seats, thus
x + y = 27000
As said, daily receipts were 204000$. As reserved seat is 13$, y of them gave 13$ each (y*13$) and x of them gave 6 for general admission(x*6) and those two add up and we get second equation
13y + 6x = 204000.
This can be solved by transforming first equation into x = 27000 - y and then replacing the x in second.
13y + 6*(27000 - y) = 204000
13y + 162000 - 6y = 204000
7y = 42000
y = 6000
x + 6000 = 27000
x = 27000 - 6000 = 21000

5 0

3 years ago

As an estimation we are told 5 miles is 8 km. <br> Convert 32.5 miles to km.

elixir [45]

Answer:

52 km

Step-by-step explanation:

Given, 5 miles is 8km.

So, 1 mile will be (8/5) km.

Now, 32.5 miles will be (8/5) x 32.5 = 52 km

5 0

3 years ago