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3 years ago

Please please please help

Mathematics

1 answer:

klio [65]3 years ago

8 0

D. 270 degrees counterclockwise
hope this help!

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PLZ ANSWER THIS QUESTION!

mafiozo [28]

17⁻¹⁸ / 17⁷=17⁻¹⁸⁻⁷=17⁻²⁵
a^n / a^m=a^(n-m)

Answer: 17⁻¹⁸ / 17⁷=17⁻²⁵

b)
11⁻⁶ / 11⁻⁷=11⁻⁶⁻(⁻⁷)=11⁻⁶⁺⁷=11¹=11

Answer: 11⁻⁶ / 11⁻⁷=11

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4 years ago

Katie ordered candles that cost $3 each from a website. She was charged a flat shipping fee of $6 and no sales tax. If she spent

vaieri [72.5K]

6c = 54 Bcz "6 She was charged a flat shipping fee of $6 and no sales tax" ;)

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4 years ago

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Which of the following linear equations has exactly one solution?

Andrei [34K]

D the answer is because

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3 years ago

What is -2x + 11 + 6x when simplifying

Vika [28.1K]

Answer:

4x + 11

Step-by-step explanation:

6x - 2x

= 4x

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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