All categories

2 years ago

3 feet of ribbons are needed for wrapping each present. How many presents can be wrapped with 102cm of ribbons?

2 answers:

8 0

1 imperial foot has approximately 30.48 metric centimeters, and for one present we need 3 ft, or namely 3(30.48) cm, how many can we get from 102 cm? 102 ÷ 3(30.48) ≈ 1.1155, so barely just one present.

kozerog [31]2 years ago

3 0

Answer:

1 present

Step-by-step explanation:

3 feet = 36 inches

1 inch = 2.54 cm, so 36 in = 36*2.54cm = 91.44cm

Each present is 91.44 cm, and 102 cm of ribbon can only be used to wrap one present.

The answer is 1 present

Hope this helps :)

Have a great day!

You might be interested in

Please help me with the question

ivanzaharov [21]

Answer:

Step-by-step explanation:

Since it's at the same time of the day, the ratio between the height of the person and the shadow they cast will stay the same. So the man's height to shadow ratio is 6:8 = 3:4. The son's height to shadow ratio would be the same so x:4 = 3:4 therefore his height is 3 feet.

4 0

2 years ago

The difference of a number N and 9 is fewer than 4

denis23 [38]

Any number less than or equal to 12

5 0

2 years ago

-0.75 7/8 and - 5/8 in order from least to greatest

VMariaS [17]

Hello there!

The answer is:
-5/8
-0.75
7/8

5 0

2 years ago

If f(1)=10 and f(n)=-5f(n-1)-2 then find the value of f(4)

Flura [38]

Answer:

$\large\boxed{f(4)=-1292}$

Step-by-step explanation:

$f(n)=-5f(n-1)-2\\\\f(1)=10\\\\f(2)=-5f(2-1)-2=-5f(1)-2\\\text{put}\ f(1)=10:\\f(2)=-5(10)-2=-50-2=-52\\\\f(3)=-5f(3-1)-2=-5f(2)-2\\\text{put}\ f(2)=-52\\f(3)=-5(-52)-2=260-2=258\\\\f(4)=-5f(4-1)-2=-5f(3)-2\\\text{put}\ f(3)=258\\f(4)=-5(258)-2=-1290-2=-1292$

4 0

2 years ago

HELP PLEASE 10 POINTS!!!!!! A country's population in 1993 was 171 million. In 1999 it was 176 million. Estimate the population

Lilit [14]

Let 1993 = time 0 = 0.

Let 1999 = time 6 = 6

Let 2012 = time 19 = 19

So, a = 171 (million). First solve for k.

176 = 171 e^k6

176/171 = e^(k*6)

ln (176/171) = 6k

k = 1/6 ln (176/171)

So, in 2012 we have: P(19) = 171 e^(19k), where k = 1/6 ln (176/171)

Hope this helped!

3 0

2 years ago