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2 years ago

What is nearest $ 8.66 to nearest $

Mathematics

2 answers:

svlad2 [7]2 years ago

7 0

Answer:

$9 I think

Step-by-step explanation:

(hhhhhhhhaahahahahshshahahhaahaaaa

Akimi4 [234]2 years ago

3 0

Answer:

$8.66 is rounded to $9

Step-by-step explanation:

When rounding a number, you round to the nearest whole number.

If the last digit is 5 or higher (5,6,7,8,9,) it rounds up

If the last digit is 4 or lower (1,2,3,4) it rounds down

Example:

$7.60 would be rounded to $8 because 6 is higher than 5, or in this case, 60 is higher than 50

$4.40 would be rounded to $4 because 4 is lower than 5, or in this case, 40 is lower than 50

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PLZ GUYS THIS IS REALLY HARD

Marina CMI [18]

lol, there are three.

Step-by-step explanation:

thanks for the points I guess.

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2 years ago

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The table below shows the cost for different numbers of tickets.

Black_prince [1.1K]

Answer:

Step-by-step explanation: It's not going up by 6 or 2, and it's not going down in numbers, so D

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2 years ago

A farmer has 520 feet of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one s

Setler [38]

Answer:

$310\text{ feet and }210\text{ feet}$

Step-by-step explanation:

GIVEN: A farmer has $520 \text{ feet}$ of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is $310 \text{ feet}$ .

TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.

SOLUTION:

Let the length of rectangle be $x$ and $y$

perimeter of rectangular pen $=2(x+y)=520\text{ feet}$

$x+y=260$

$y=260-x$

area of rectangular pen $=\text{length}\times\text{width}$

$=xy$

putting value of $y$

$=x(260-x)$

$=260x-x^2$

to maximize $\frac{d \text{(area)}}{dx}=0$

$260-2x=0$

$x=130\text{ feet}$

$y=390\text{ feet}$

but the dimensions must be lesser or equal to than that of barn.

therefore maximum length rectangular pen $=310\text{ feet}$

width of rectangular pen $=210\text{ feet}$

Maximum area of rectangular pen $=310\times210=65100\text{ feet}^2$

Hence maximum area of rectangular pen is $65100\text{ feet}^2$ and dimensions are $310\text{ feet and }210\text{ feet}$

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2 years ago

What is the sum of 7/9 plus 3/5

Ivan

Answer:

$\frac{62}{45}$

Step-by-step explanation:

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2 years ago

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago