The total area of the shape is 32

<h3><u>Analysis </u><u>of </u><u>graph </u><u>:</u><u>-</u></h3>
We have given one graph which is plot between distance and time. where time is in minutes and distance is in seconds.
<u>According </u><u>to </u><u>the </u><u>graph </u>
- For the first 5 minute ( O to A) , The distance is continously increasing 2m / per minute .
- For the 5 minute that is from 5 minute to 13 minute ( A to B) both marellize and her dog wally moving with the constant speed .
- For next 3 minutes that is from 10 minutes to 13 minutes ( B to C) , The distance is continously decreasing with time .
- For next 3 minutes that is from 13 to 16 minutes ( C to D) , Again they moved with constant speed .
- For next 6 minute that is from 16 to 21 minutes ( D to E) . Again, There distance is increasing with time .
- Again For next 4 minutes that is 21 to 25 minutes , they are moving with constant velocity .
<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>
1) Between O and A
- The marellize and wally when moving between O to A , The distance is constantly increasing with time.
- The graph is Straight line
2) Between A and B
- The marellize and wally when moving between A to B, The distance remains the same with time that is they moving with constant speed.
- The graph is constant or steady
3) Between B to C
- The marellize and wally when moving between B to C, The distance is constantly decreasing with time .
- The graph is straight line but it follows decreasing function .
4) For covering the First 6 km ,
<u>According </u><u>to </u><u>the </u><u>graph</u><u>, </u>
- For covering first 6 km, They took 3 minutes.
5) No, Marellize and wally walk does not from where they have started.
<u>According </u><u>to </u><u>the </u><u>graph </u>
- It is end at 5 m instead of 0m .
Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
We are required to simplify the quotient: ![\dfrac{\sqrt[3]{60} }{\sqrt[3]{20}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B60%7D%20%7D%7B%5Csqrt%5B3%5D%7B20%7D%7D)
Since the <u>numerator and denominator both have the same root index</u>, we can therefore say:
![\dfrac{\sqrt[3]{60} }{\sqrt[3]{20}} =\sqrt[3]{\dfrac{60} {20}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B60%7D%20%7D%7B%5Csqrt%5B3%5D%7B20%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B60%7D%20%7B20%7D%7D)
![=\sqrt[3]{3}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B3%7D)
The simplified form of the given quotient is
.
6>4-2x<4
so therefor (4x-2) must be less than 6 and less than 4
it must be
6>4-2x and 4<4-2x
find the intersection
6>4-2x
add 2x to both sides
6+2x>4
subtract 6
2x>-2
divide by 2
x>-1
4>4-2x
add 2x
4+2x>4
subtract 4
2x>0
x>0
so we have
x>0
and x>-1
the range of x>-1 includes most of x>0 so the answer is
x>-1
Jewelery maker has
24 jade beads
30 teak Beads
So we need to find the highest common factor for both 24 and 30
The factors for both numbers are
24 -1,2,3,4,6,8,12,24
30 - 1,2,3,5,6,10,15,30
The highest common factor for both is 6
The greatest number of necklaces she can make is 6
Number of jade beads - 24/6 = 4 jade beads
Number of teak beads - 30/6 = 5 teak beads
So each necklace will have 4 jade beads and 5 teak beads