Write an inequality for the sentence. c is not less than zero.

HURRY I NEED HELP: Write an equation in point-slope form of the line that passes through the point (3, 5) and has a slope of m=−

skelet666 [1.2K]

Answer:

$\huge y - 5 = - 1(x - 3) \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y - 5 = - 1(x - 3)$

Hope this helps you

6 0

3 years ago

10. For f(x) = 3 - 5x, find f(-5). A. –22 B. -7 C. 3 D. 28

Artist 52 [7]

Answer:

D

Step-by-step explanation:

$3 - 5( - 5) \\ 3 - ( - 25) \\ 3 + 25 \\ 28$

4 0

3 years ago

An ice cream shop sold 48 vanilla milkshakes in a day, which was 40% of the total number of milkshakes sold that day. What was t

german

Answer:

They sold 120 milkshakes that day

Step-by-step explanation:

We can set up a proportion

$\frac{48}{m} =\frac{2}{5}$

$48*5=2m\\ 2m = 240\\m = 120\\$

Note: If you have some value, $v$ , and it is $\frac{x}{100}$ of the total amount, you can find the value by doing the equation $\frac{v}{x}*100$

8 0

4 years ago

Help me show work please

skad [1K]

X+X+3xX+4 divided by X = F

5 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago