Answer:
See Explanation
Step-by-step explanation:
Given

Required
Determine the weights that approximates to 41.3kg
The options are not given but the question is still solvable
In approximation:
0 - 4 approximates to 0
This implies that:
41.30 to 41.34 approximates to 41.3
Also:
5 - 9 approximates to 1
This implies that:
41.25 to 41.29 approximates to 41.3
Bring the range together, we have:

<em>Select any three numbers within this range, and you have your answer</em>
Examples are: <em>41.25, 41.28, 41.34</em>\
Answer:
2\10
Step-by-step explanation:
1\2 = 5\10
7\10 - 2\10 = 5\10
Step-by-step explanation:
<h3><u>QU</u><u>ESTION</u><u>:</u><u>-</u></h3>
<u>Fin</u>d 26th term of the A.P --15,-35,-55,-75...
<h3><u>Solu</u><u>tion</u><u>:</u><u>-</u></h3>
First term=a=-15
Common Difference=d=-20
We know that






Answer:
The width is 20 ft and the length is 80 ft
Step-by-step explanation:
Area = l * w
l = w +60
SA = 1600 ft^2 ( which is the regular area) The surface area of the water is just what is on top which is length times width
1600 = l*w
Substitute in l = w+60
1600 = (w+60) *w
Distribute the w
1600 = w^2 +60w
Subtract 1600 from each side
1600-1600 = w^2 + 60w -1600
0 = w^2 + 60w -1600
Factor
0=(w-20) (w+80)
Using the zero product property
w-20=0 w+80 =0
w=20 or w=-80
The width cannot be negative so
w=20
Now we need to find the length
l = w+60
l = 60+20
l =80
Hello there!
Math is hard! So we as students, we need to play the smart game with Math in order to find the easiest way to solve the problems. What do I mean? Well as you can see they gave us Fraction which won't be easy! So we will transform the fractions into decimals. So let's get started.
Then a scale of 1/4. Transformed into decimal = Then a scale of 0.25
So instead of using the fraction 1/4, I used its decimal 0.25
So if:
48 feet ......> 12 inches
x feet ........> 0.25 inches
Cross multiply
12 * x = 0.25 * 48
12x = 12
Divide both sides by 12
12x/12 = 12/12
x = 1
The correct answer is option D, 1 foot
I hope this helps! Please let me know if you have any question.
Thanks!