Solving this requires use of conversions
we already know that
2.05 m > 2 m
all that's left is the question
200 cm vs. 2.05 meters
I we know our equivalencies, we know that
100 cm = 1 m
if this is so then
200 cm = 2 m
we already know that
2.05 m > 2 m
so it must also be that
2.05 m > 200 cm
Answer:
- perimter of original rectangle = <u>17. 6 mm</u>
- side length of the enlarged rectangle = <u>23. 22 mm</u>
- perimeter of the enlarged rectangle = <u>95. 04 mm</u>
Step-by-step explanation:
<u>PERIMETER</u><u> </u><u>OF</u><u> </u><u>ORIGINAL</u><u> </u><u>RECTANGLE</u>
- Length of original rectangle = 4.5 mm
- Width of original rectangle = 4.3 mm
<em>perimeter = 2 × ( length + width)</em>
= 2 × ( 4.5 + 4.3)
= 2 × 8.8
= 17. 6 mm
<u>SIDE</u><u> </u><u>LENGTH</u><u> </u><u>OF</u><u> </u><u>ENLARGED</u><u> </u><u>RECTANGLE</u>
- Width of original rectangle = 4. 5 mm
- Width of enlarged rectangle = 24.3 mm
Enlargement factor = 24.3 / 4.5
= 5.4
- Length of original rectangle = 4.5 mm
- Enlargement factor = 5.4
Side length of enlarged rectangle
= original length × Enlargement factor
= 4.3 × 5.4
= 23. 22 mm
<u>PERIMTER OF ENLARGED RECTANGLE</u>
= 2 × ( enlarged ength + enlarged breadth)
= 2 × (23. 22 + 24. 3 )
= 95. 04 mm
Answer:
The first one
Step-by-step explanation:
An easy way is SOH CAH TOA
SOH is that you use sine if you there is the opposite (in this case the question mark) and hypotenuse (the diagonal line)
CAH is that you use cosine for the adjacent side (1100) and the hypotenuse.
TOA is using tangent for the opposite and adjacent side.
To fine the question mark, you would use tangent. And plug in the given (1100 for the adjacent). So tan45 = x/1100.
If you need more help or clarification, let me know!
Adding the numbers gives us a total of 65
Jan = 20/65 = 30.7% = 31%
Feb = 25/65 = 38.4% = 38%
Mar = 1/65 = 1.5% = 2%
Apr = 3/65 = 4.6% = 5%
May = 16/65 = 24.6% = 25%
I believe ur answer is 30%
To solve this problem, we can use the tan function to find
for the distances covered.
tan θ = o / a
Where,
θ = angle = 90° - angle of depression
o = side opposite to the angle = distance of boat from
lighthouse
a = side adjacent to the angle = height of lighthouse = 200
ft
When the angle of depression is 16°18', the initial distance
from the lighthouse is:
o = 200 tan (90° - 16°18')
o = 683.95 ft
When the angle of depression is 48°51', the final distance
from the lighthouse is:
o = 200 tan (90° - 48°51')
o = 174.78 ft
Therefore the total distance the boat travelled is:
d = 683.95 ft - 174.78 ft
<span>d = 509.17
ft</span>