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3 years ago

7

The following equation has been solved incorrectly. Explain the students error

2 answers:

balandron [24]3 years ago

6 0

On the right side of the equation, when the student combined like terms, 7x and -4x, they did not combine them correctly. Instead of combining them correctly, the student forgot to include the sign of the -4x which caused the sum of the two terms to be 11x instead of 3x. Hope this helps!

svlad2 [7]3 years ago

3 0

What he said ^^ no problem

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If f(x) = 3x2 − 8x, 0 ≤ x ≤ 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints.

denis23 [38]

Split up the interval [0, 3] into 6 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2], [2, 5/2], [5/2, 3]

The right endpoints are given by the arithmetic sequence,

$r_i=0+\dfrac i2=\dfrac i2$

with $1\le i\le6$ .

We approximate the integral of $f(x)$ on the interval [0, 3] by the Riemann sum,

$\displaystyle\int_0^3f(x)\,\mathrm dx=\sum_{i=1}^6f(r_i)\Delta x_i$

$\displaystyle=\frac{3-0}6\sum_{i=1}^6\left(3{r_i}^2-8r_i\right)$

$\displaystyle=\frac12\sum_{i=1}^6\left(\frac{3i^2}4-4i\right)$

$\displaystyle=\frac38\sum_{i=1}^6i^2-2\sum_{i=1}^6i$

Recall the formulas,

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

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

Then the value of the integral is approximately

$\displaystyle=\frac38\cdot\frac{6\cdot7\cdot13}6-2\cdot\frac{6\cdot7}2=\boxed{-\frac{63}8}=-7.875$

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7 0

3 years ago

Help quick please. Order the numbers from least to greatest.

Anastasy [175]

Answer:

$-2\frac{1}{4}$

Step-by-step explanation:

Mod of any number represents the absolute value of the number.

Therefore, $|-2\frac{1}{4}|=2\frac{1}{4}$ = 2.25

$-2\frac{1}{4}=-2.25$

$|1\frac{1}{4}|=1.25$

$\frac{3}{4}=0.75$

$|-1\frac{3}{4}|=1\frac{3}{4}=1.75$

$-1\frac{1}{4}=-1.25$

Now we can arrange these numbers in ascending order.

-2.25 < -1.25 < 0.75 < 1.25 < 1.75 < 2.25

Therefore, $-2\frac{1}{4}$

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olga_2 [115]

Answer:

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Step-by-step explanation:

The line starts at 2, and is going up by a constant rate or 2/1.

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