All categories

2 years ago

Need work for problems I already know the answers

Mathematics

1 answer:

alexgriva [62]2 years ago

4 0

Don’t take this comment seriously

You might be interested in

Simplify -3d^8(-4d^-14). Assume d =/ 0

Alenkinab [10]

Answer:

$\frac{12}{d^6}$

Step-by-step explanation:

$-3d^8\left(-4d^{-14}\right)$

Double negative = Positive:

$3d^8*4d^{-14}$

Apply negative exponent formula:

$3*d^8*4*\frac{1}{d^{14}}$

Multiply:

$\frac{12d^8}{d^{14}}$

Apply exponent rule: $\frac{x^a}{x^b}=\frac{1}{x^{b-a}}$

Thus, $\frac{12}{d^{14-8}}$

Simplify:

$\frac{12}{d^6}$

8 0

3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval (0, 3) into n equal subinterv

Viktor [21]

Splitting up [0, 3] into $n$ equally-spaced subintervals of length $\Delta x=\frac{3-0}n = \frac3n$ gives the partition

$\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]$

where the right endpoint of the $i$ -th subinterval is given by the sequence

$r_i = \dfrac{3i}n$

for $i\in\{1,2,3,\ldots,n\}$ .

Then the definite integral is given by the infinite Riemann sum

$\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}$

8 0

2 years ago

Model factors: use tiles to find all the factors of the product, record the arrays on grid paper and write the factors shown.

Ipatiy [6.2K]

The factors is just 1 and 23. The array should be 23 rows/columns of just 1 dot only. Here's an example:

Factors of 5: 1 and 5

Array: •••••

7 0

3 years ago

What is the value of b?

VikaD [51]

Answer: The b-value is the middle number, the number next to the X

5 0

3 years ago

Every even number,except 2,is composite

Degger [83]

Yes. That's correct. Is that what you're asking...? To clarify, prime numbers have a single factor (one and itself) , composite numbers have multiple factors.

5 0

4 years ago