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3 years ago

Wha is 3/8×5/6 you will need a paper.

Mathematics

2 answers:

IRISSAK [1]3 years ago

8 0

Answer:

Hey There!

The solution is

5/16 or 0.3125

icang [17]3 years ago

4 0

3/8 x 5/6 = 15/48 = 5/16

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An underestimate of 532 times 11

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An underestimate of 532 times 11 is 5852

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3 years ago

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What is the product of 0.42 and 0.03?

ArbitrLikvidat [17]

The product is 0.0126

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3 years ago

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A corporate team-building event costs $19plus an additional $1 per attendee. How many attendees can there be, at most, if the bu

Mars2501 [29]

Answer:

There can be at most $12$ attendees in a corporate team-building event.

Step-by-step explanation:

Let $x$ denotes number of attendees in a corporate team-building event.

Fixed cost = $19

Cost charged per attendee = $1

Budget for the corporate team-building event = $31

Therefore,

$19+1(x)\leq 31\\19+x\leq 31\\x\leq 31-19\\x\leq 12$

So, there can be at most $12$ attendees in a corporate team-building event.

8 0

3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

3 0

1 year ago

Consider the perfect square trinomial identity:

denis-greek [22]

Answer:

a = x

b = 5

Step-by-step explanation:

5 0

4 years ago