Ask question

Ask question

All categories

faltersainse [42]

2 years ago

8

What is the square root of 460

2 answers:

Volgvan2 years ago

8 0

Answer:

21.4476105895

Step-by-step explanation:

Using a calculator $\sqrt{460\\}$

Dmitry [639]2 years ago

6 0

Answer:

approximately 21.448 or 2 times the square root of 115

Step-by-step explanation:

You might be interested in

A, B, and C are collinear, and B is between A and C. The ratio of AB to BC is 1:4.

schepotkina [342]

Let point C has coordinates $(x_C,y_C).$ Consider vectors

$\overrightarrow{AB}=(x_B-x_A,y_B-y_A)=(8-9,6-9)=(-1,-3),\\ \\\overrightarrow{BC}=(x_C-x_B,y_C-y_B)=(x_C-8,y_C-6).$

Since the ratio AB to BC is 1:4, you have that

$\dfrac{-1}{x_C-8}=\dfrac{1}{4}\quad \text{and}\quad \dfrac{-3}{y_C-6}=\dfrac{1}{4}.$

Find $x_C$ and $y_C:$

$x_C-8=-4,\\ \\x_C=-4+8=4,\\ \\y_C-6=-3\cdot 4=-12,\\ \\y_C=-12+6=-6.$

Answer: C(4,-6)

7 0

3 years ago

Read 2 more answers

What is the length, width, and area

Temka [501]

Length: 6 Width: 2 Area: 12

4 0

3 years ago

What is the slope of the line that passes through (1,-4) and (-4,-6)

Fudgin [204]

The slope of the line is 4/10

remember: rise over run

6 0

2 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

you are to give a client one tablet labeled 0.15 mg and one labeled 0.025 mg. what is the total dosage of these two tablets

Ivenika [448]

Simply add the two dosages together, (0.15+0.025) and the answer is 0.175 :)

7 0

3 years ago