Help please.........

What sign is it > or < plzz help :((((((:((

andrew11 [14]

Answer:

c<1

Step-by-step explanation:

Less than 1.

3 0

3 years ago

The graph of F(x), shown below, has the same shape as the graph of

dimaraw [331]

<em>Greetings from Brasil...</em>

<em />

We know that the translations are established as follows:

→ Horizontal

F(X + k) ⇒ k units to the left

F(X - k) ⇒ k units to the right

→ Vertical

F(X) + k ⇒ k units up

F(X) - k ⇒ k units down

G(X) = X and F(X) is G(X) shifted up 2 units.

In the statement it is said that there was a translation of 2 units upwards, so

F(X) = G(X) + k where k = 2 units up

3 0

3 years ago

Solve the inequality and express your answer in interval notation.<br><br> -11 <-37 +4 <-8

Grace [21]

Answer:

(-11,-8)

Step-by-step explanation:

7 0

3 years ago

Zach is 4 years older than twice his sister Maya’s age. Let m represent Maya’s age.

rewona [7]

C. because 2m is twice maya’s age and the +4 is the four years older than twice, making it 2m+4

3 0

3 years ago

I need help with 2,3 and 6! Please Help!

Georgia [21]

2. If you already know Faulhaber's formula, which says

$\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6$

then it's just a matter of setting $n=4$ . If you don't, then you can prove that it works (via induction), or compute the sum by some other means. Presumably you're not expected to use brute force and just add the squares of 1 through 4.

Just to demonstrate one possible method of verifying the formula, suppose we start from the binomial expansion of $(k-1)^3$ , do some manipulation, then sum over $1\le k\le n$ :

$(k-1)^3=k^3-3k^2+3k-1$
$\implies k^3-(k-1)^3=3k^2-3k+1$
$\implies\displaystyle\sum_{k=1}^n(k^3-(k-1)^3)=\sum_{k=1}^n(3k^2-3k+1)$

The left side is a telescoping series - several terms in consecutive terms of the series will cancel - and reduces to $n^3$ . For example,

$\displaystyle\sum_{k=3}^2(k^3-(k-1)^3)=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)=3^3$

Distributing the sum on the right side across each term and pull out constant factors to get

$\displaystyle n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1$

If you don't know the formula for $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$ , you can use a similar trick with the binomial expansion $(k-1)^2$ , or a simpler trick due to Gauss, or other methods. I'll assume you know it to save space for the other parts of your question. We then have

$\displaystyle n^3=3\sum_{k=1}^nk^2-\frac{3n(n+1)}2+n$
$\implies\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6$

and when $n=4$ we get 30.

3. Each term in the sum is a cube, but the sign changes. Recall that $(-1)^n$ is either 1 if $n$ is even or -1 if $n$ is odd. So we can write

$1^3-2^3+3^3-4^3+5^3=\displaystyle\sum_{k=1}^5(-1)^{k-1}k^3$

( $k+1$ as the exponent to -1 also works)

6. If $0\le k\le n$ , and $i=k+1$ , then we would get $1\le k+1\le n+1\iff1\le i\le n+1$ . So the sum with respect to $i$ is

$\displaystyle\sum_{k=0}^n\frac{k^2}{k+n}=\sum_{i=1}^{n+1}\frac{(i-1)^2}{i+n-1}$

4 0

3 years ago