The arca of a picture 25 7/5 square inches.If the width of the picture is 4 1/2 inches.

Mathematics

1 answer:

Stella [2.4K]3 years ago

8 0

Answer:

6 in

Step-by-step explanation:

The area of a picture is 25 7/5 square inches. If the width of the picture is 4 1/2 inches, what is the length? (A=(w)(l) )

Assuming that your "25 7/5 square inches" is not a mistake, then this is equivalent to 26 2/5 square inches.

Converting 26 2/5 sq. in. to an improper fraction, we get 132/5 sq. in., which represents the area of the picture. If the width is 4 1/2 inches, or 9/2 inches, then the length, l, is:

132/5 in^2

l = ------------------

9/2 in

To divide 132/5 in^2 by the fraction 9/2 in, invert 9/2 and multiply:

135 in^2 2

l = -------------- * ------ = 270/9 in = 30/5 in = 6 in

5 9

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Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six subintervals, taking the sample points, xi, to be the right e

DENIUS [597]

Hello,

Reminders:
$$\sum_{i=1}^{n} i= \dfrac{n(n+1)}{2}$


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

$
$$ \sum_{i=1}^{n} i^3= \dfrac{n^2(n+1)^2}{4} $
$

$n=6\\
x_{0}=0\\
x_{n}=3\\
\Delta= \dfrac{x_n-x_0}{n}=0.5\\

$
$x_{i} =0+\Delta*i= \frac{i}{2}$

$\sum_{i=1}^{n}\ \Delta*f(x_{i})=\sum_{i=1}^{6}\ \dfrac{i^3/8-6i}{2}$
$$= \dfrac{1}{2}\sum_{i=1}^{6} (\frac{i^3}{8} -6i)=\dfrac{1}{16}*\frac{(6*7)^2}{4}- \frac{9*7}{2}= -3.9575$

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