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3 years ago

The limit as h approaches 0 of (e^(2+h)-e^2)/h = ?

1 answer:

6 0

First, separate the exponents of the first exponential:

$\lim_{h \to 0} \frac{e^{2} e^{h} - e^{2} }{h}$

Regroup e² on the numerator:
$\lim_{h \to 0} \frac{ e^{2} (e^{h} - 1) }{h}$

e² is now a constant that can be brought outside the limit:

$e^{2} \lim_{h \to 0} \frac{ e^{h} - 1}{h} 
$

This limit is a notable one, and we know it is equal to 1,

therefore:

e² · 1 = e²

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Q.The rectangular front windshield of the bus has the length of 1.5√3 m and the width of 1.5 m . The wipers are attached to the

galina1969 [7]

Answer:

Area covered by wiper=area of arc(A0D)+area of traingleAOB+area of arc(BOC)

We know AO and OB are wipers,whose lengths are 1.5

In ∆ABC

AB^2+BC^2=AC^2

(1.5√3)^2+(1.5)^2=(AC)^2

2.25×3+2.25=(AC)^2

2.25×4=AC^2

AC=3

Also AO=1.5

similarly using Pythagoras theorem in ∆OBC

we get

OC=1.5

this shows that point O is the bisector of diagonal AC

so,AO=OC

also from property of rectangle we know that, diagonal bisects each other equally

so,O will also be the bisector of diagonal BD

hence OB=OD=1.5

AC=BD (property of triangle)

AC/2=BD/2

OB=OC,hence OB=OC=OD=OA=1.5=AD=BC

Hence OAD will be an equilateral triangle

angle OAD=60

similarly angleOBC=60

area of sector=angle of that sector ×area of circle/360

area of arc(AOD)+arc(BOC)=60/360 ×3.14 ×1.5×1.5 +60/360 ×3.14×1.5×1.5

=2×(60/360)×1.5×1.5

area of triangleAOB=area of rectangle/4 (diagonal divides the rectangle in 4 equal parts)

area of triangle=1.5×1.5√3/4

=>1.5×2.6/4=3.90/4

=0.975 or 0.98(approx)

Area covered by wiper=.98+2×60/360 ×3.14×1.5×1.5

=.98+0.75×3.14=2.35+0.97

=2.35+.98

Percentage of area of windshield covered by wiper=3.3×100/3.9=84.61

hence percentage of area of windshield covered by wiper is 85percent(approx)

plz mark brainlest

5 0

3 years ago

Francine Flicka found a bargain when she bought a lawn mower. She paid the store's clerk three $20 bills and received him $7.45

adoni [48]

Answer:

$48.97

Step-by-step explanation:

Let p represent the marked price of the mower. Then the price with tax is ...

with tax = p + 7.3%×p = 1.073p

The amount Flicka paid is the amount tendered less the change she received, so is ...

3×$20 - 7.45 = with tax = 1.073p

To find p, we can divide by its coefficient:

(3 × $20 - 7.45)/1.073 = p ≈ $48.97

The selling price before taxes was $48.97.

_____

<em>Comment on the problem</em>

If you check the answer, you find the amount with tax is ...

$48.97 × 1.073 ≈ $52.54481 ≈ $52.54

so the change Flicka would have received would have been $7.46, not $7.45. If the price were $48.98, then the change would have been $7.44. There is no price at which the mower can be marked that will make the total with tax come to $52.55. There is no solution to this problem.