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3 years ago

5

Jim wants to put a wallpaper border around a rectangular room.The room measures 14 feet by 10 feet. If the wallpaper border come

s in stripes that are 20 feet long, how many strips will Jim need?

1 answer:

Marat540 [252]3 years ago

3 0

Answer: He needs 3 stripes.

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2 ÷ 48.9 plss need help

Mkey [24]

Answer:

0.04

I rounded to one significant figures.

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3 years ago

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is 14.8 nm . What is the total wi

IrinaK [193]

Answer: 1539.2nm

The space between molecule is going as X-1 function. That mean 1 molecule doesn't have space, 2 molecules have 1 space, 3 molecules have 2 space. Then 105 molecules will be will have 104 space.
Each space is 14.8nm, the total distance would be: 104 space x 14.8 nm/space= 1539.2nm

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3 years ago

WILL GIVE BRAINLIEST IF CORRECT!!<br> HELP PLS!!

bulgar [2K]

Answer:

the answer should be B

Step-by-step explanation:

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3 years ago

Select the correct answer from each drop-down menu.

Oksana_A [137]

Answer:

B

Step-by-step explanation:

B because range of set B is more than set A

range B=56-12=44

range A=55-12=43

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2 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago