P and q are prime numbers. p3 x q = 56. Find p and q.

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

6 0

2 years ago

Here is the histogram of a data distribution. What is the shape of this distribution?

Pavlova-9 [17]

Answer:

Unimodal-Skewed

Step-by-step explanation:

A distribution is called unimodal if it has only one hump in the histogram.

A symmetric distribution is equally divided on both sides of the highest hump.

The given histogram has only one hump at 4 and as it is not symmetrically distributed, it is skewed.

So the correct answer is:

Unimodal-Skewed ..

7 0

2 years ago

Holly checked her answer to a division problem by estimating the quotient. Which is the best estimation of the quotient?

Andre45 [30]

Answer: 4

Try just dividing 23/5 and you’d get 4 something. But since 5.97 is close to 6 you can also mentally divide 23/6 and get 3something (it’d be really close to 4)

5 0

2 years ago

Read 2 more answers

Define fn : [0,1] --> R by the

sasho [114]

Answer:

The sequence of functions $\{x^{n}\}_{n\in \mathbb{N}}$ converges to the function

$f(x)=\begin{cases}0&0\leq x$ .

Step-by-step explanation:

The limit $\lim_{n\to \infty }c^{n}$ exists and converges to zero whenever $\lvert c \rvert$ . But, if $c=1$ the sequence $\{c^{n}\}$ is constant and all its terms are equal to $1$ , then converges to $1$ . Using this result, consider the sequence of functions $\{f_{n}\}$ defined on the interval $[0,1]$ by $f_{n}(x)=x^{n}$ . Then, for all $0\leq x$ we have that $\lim_{n\to \infty}x^{n}=0$ . Now, if $x=1$ , then $\lim_{n\to \infty }x^{n}=1$ . Therefore, the limit function of the sequence of functions is