The area of the following rectangle is 6O and^4 square meters and its length is 30 and meters

Length=x
width=y
Perimeter of a rectangle=2(length)+2(width)
Therefore:
2x+2y=200
We simplify the equation dividend both sides of this equation by 2:
x+y=100
Then:
y=100-x

Area of a rectangle: length x width
We have the next inequation:
x(100-x)<900
100x-x²<900
x²-100x+900<0

We solve this inequation
1) we solve this equation:
x²-100x+900=0
x=[100⁺₋√(10000-3600)]/2=(100⁺₋80)/2
We have two solutions in this equation:
x₁=90
x₂=10

2)With these values, we make intervals:
(-∞,10)
(10,90)
(90,∞)

3)With these intervals, we check it out if the inequation works:
(-∞,10); for example; if x=0 ⇒ 0²-100(0)+900=100>0, this interval don´t work.
(10,90);f.e: if x=11; ⇒ 11²-100(11)+900=-79<0, this interval works.
(90,∞); fe; if x=91 ⇒91²-(100)91+900=81>0; this interval don´t work.

answer: the possible lengths would be the values inside of this interval:
(10,90) ft.

7 0

4 years ago

Andra travels 90 miles at an average speed of 70 mph.

spayn [35]

Answer:

54mph

Step-by-step explanation:

$\textsf {\underline {To find :-}}$

The average speed at 2nd departure.

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$\textsf{ \underline {Given :-}}$

The speed during 90 miles travel = 70mph

The average speed during the entire journey = 63mph

The distance in second departure = 70 miles

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$\textsf {\large{ \underline{ Solution :-}}}$

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The distance travelled in the entire journey

= 90 + 70 = 160 miles.

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Then the time taken in entire journey will be

$\sf \large = \frac{total \: distance}{average \: speed} \\ \\ \sf \large = \frac{160}{63} \\ \\ \sf \large = 2 \frac{37}{63}hours \\ \\ \sf \large = 2 \: hours \: 35 \: minutes$

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Time taken at first departure

$\sf \large = \frac{distance}{speed} \\ \\ \sf \large = \frac{9 \cancel0 }{7 \cancel0} \\ \\ \sf \large = 1 \frac{2}{7} \: hours \\ \\ \sf \large = 1 \: hour \: 17 \: minutes$

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Time taken in second departure

= 2 hours 35 minutes - 1 hour 17 minutes

= 1 hour 18 minutes

⠀

Speed at second departure will be

$= \sf \large \frac{distance}{time} \\ \\ \sf \large = \frac{70}{1 \frac{18}{60} } \\ \\ \sf \large = \frac{70}{1.3} \\ \\ \sf \large = 53.84 \: mph \\ \\ \sf \large = 54 \: mph$