All categories

3 years ago

Help me fast please

Mathematics

1 answer:

solong [7]3 years ago

3 0

Answer:

81.57 cm

Step-by-step explanation:

divide the actual length by 100 =27.19

times 27.19 by 3= 81.57 cm hope this helps, have a nice day :)

You might be interested in

HELP MEEE!! Please evaluating a compound. Inequality using a graph

Irina-Kira [14]

Answer:

$\frac{x}{2}$

Step-by-step explanation:

From the graph, we can conclude that,

$x$ is less than 2 and not including as there is a hollow circle at the mark 2.

Also, $x$ is greater than or equal to 7 including 7 as there is a solid circle at the mark 7

So, the compound inequality will be $x$

Now, the option that simplifies to the above inequality is the required answer.

Let us check the first option.

$\frac{x}{2}$

Therefore, option 1 simplifies to the above compound inequality.

So, the correct answer is option 1.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%28%5Cfrac%7B8a%5E%7B-5%7Db%5E%7B4%7D%20%7D%7B12a%5E%7B-6%7Db%5E%7B-2%7D%20%7D%29" id="TexForm

Oxana [17]

Answer:

$→ \frac{8 {a}^{( - 5) } {b}^{(4)} }{12 {a}^{( - 6)} {b}^{( - 2)} } = \frac{2 {a}^{(6 - 5)}{b}^{(4 + 2)} }{3} = \boxed{ \frac{2}{3}a {b}^{6} }✓ \\$

<u>(2/3)ab⁶</u> is the right answer.

6 0

3 years ago

What is the sum of the whole number from 1 to 2000

lidiya [134]

Since it said sum it might be 1 times 200 so 2000

5 0

3 years ago

Read 2 more answers

Given the table of values below, which of the following ordered pairs are found on the graph of the inverse function?

balu736 [363]

Answer:

Step-by-step explanation:

7 0

3 years ago

Use the backward substitution method to obtain the closed form formula for the recurrence relation

abruzzese [7]

Answer:

$a_0=0; \; a_n=2^n-1 \; (n>0)$

Step-by-step explanation:

$a_0=0$

$a_1=2a_0+1=1$

$a_2=2a_1+1=3=1+2$

$a_3=2a_2+1=7=1+2+4$

$a_4=2a_3+1=15=1+2+4+8$

$a_5=2a_4+1=31=1+2+4+8+16$

So, we can infer that

$a_n=1+2^1+2^2+2^3+...+2^{n-1}$

and we only need to find out a formula for this sum (a geometric reason with common ratio 2)

Let's call

$S=1+2^1+2^2+...2^{n-1}$

then,

$2S=2+2^2+...2^{n-1}+2^n$

hence,

$S-2S=1-2^n$

$2S-S=S=2^n-1$

and we have our closed formula for the sequence

$a_0=0; \; a_n=2^n-1 \; (n>0)$

3 0

3 years ago