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3 years ago

Calculate the area of the following rectangle if a= 11.9 and b= 6.6

Mathematics

1 answer:

Sladkaya [172]3 years ago

7 0

we know that

area of rectangle = length*width

we are given

length=a=11.9

width=b=6.6

$Area=a*b$

now, we can plug values

and we get

$Area=11.9*6.6$

$Area=78.54$ ........Answer

so, option-C.......Answer

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Which number produces a rational number when added to 0.25

mina [271]

Answer:

$C)~ \dfrac 29$

Step-by-step explanation:

$\text{The sum of two rational numbers is rational.}$

7 0

2 years ago

Use the form of the definition of the integral given in the theorem to evaluate the integral. ∫ 0 − 2 ( 7 x 2 + 7 x ) d x

Murrr4er [49]

Answer:

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

Step-by-step explanation:

The definite integral of a continuous function <em>f</em> over the interval [a,b] denoted by $\int\limits^b_a {f(x)} \, dx$ , is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,

$\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty} \sum_{i=1}^{n}\Delta x \cdot f(x_i)$

where $\Delta x = \frac{b-a}{n}$ and $x_i=a+\Delta x\cdot i$

To evaluate the integral

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx$

you must:

Find $\Delta x$

$\Delta x = \frac{b-a}{n}=\frac{0+2}{n}=\frac{2}{n}$

Find $x_i$

$x_i=a+\Delta x\cdot i\\x_i=-2+\frac{2i}{n}$

Therefore,

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} f(-2+\frac{2i}{n})$

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx=\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})$

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})$ $\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} 4-\frac{8i}{n}+\frac{4i^2}{n^2} -2+\frac{2i}{n}\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2$

$\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]$

We can use the facts that

$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}$

$\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(\frac{n^2-3n+2}{n^2}\right)\\\\\mathrm{Divide\:by\:highest\:denominator\:power:}\:1-\frac{3}{n}+\frac{2}{n^2}\\\\\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)\\\\\frac{14}{3}\left(\lim _{n\to \infty \:}\left(1\right)-\lim _{n\to \infty \:}\left(\frac{3}{n}\right)+\lim _{n\to \infty \:}\left(\frac{2}{n^2}\right)\right)\\\\\frac{14}{3}\left(1-0+0\right)\\\\\frac{14}{3}$