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JulsSmile [24]
4 years ago
12

While in europe, if you drive 107 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite

r and your car's gas mileage is 42.0 mi/gal ? assume that 1euro=1.26dollars?
Mathematics
2 answers:
nevsk [136]4 years ago
8 0

Answer:

The money spend on gas in one week = 46.13 Euro

The money spend on gas in one week = $58.12

Step-by-step explanation:

1 week = 7 days

You drive 107 km per day.

Kilometer driven for 7 days = 7 *107 = 749 kilometers.

Now let's convert kilometers to miles.

1 mile = 1.60934 kilometers

We have to divide 749 km by 1.60934.

So, 749 km = 465.41 miles.

Car mileage is 42 miles / gallon.

We need to find the number of gallons of gas needed to travel 465.41 miles.

So, we have to divide 465.41 miles by 42.

The number of gallons of gasoline needed = \frac{465.41}{42} = 11.08 gallons

cost of 1 .10 Euros per liter.

Now we have to convert 11.08 gallons to liters.

1 gallon = 3.78541 liters.

So we have to multiply 11.08 by 3.78541

11.08 gallons = 41.94 liters.

Now to find the cost, we need to multiply 41.94 liters by 1.10 Euro.

The money spend on gas in one week = 46.13 Euro

To convert the cost in dollars.

1 euro = 1.26

We need to multiply 46.13 by 1.26

The money spend on gas in one week = $58.12

Tcecarenko [31]4 years ago
5 0
107 km is 65 miles, 4 liters is 1 gallon so if you went 42 miles then you spent like 3 or 4 dollar s
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Gre4nikov [31]
<h3><u>Answer:</u></h3>

\large\boxed{\pink{\sf \leadsto Value \ of \ LM \ is \ 11 \ units . }}

<h3><u>Step-by-step explanation:</u></h3>

A figure is given to us in which we can see two triangles one is ∆ MPL and other is ∆MPN .

<u>Figure</u><u> </u><u>:</u><u>-</u><u> </u>

\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\linethickness{0.25mm}\put(0,0){\line(1,2){2}} \put(0.001,0){\line(1,0){4}}\put(2,0){\vector(0,1){5}}\put(4,0){\line( - 1,2){2}}\put(0,-0.4){$\bf L $}\put(2,-0.4){$\bf P$}\put(4,-0.4){$\bf N $}\put(2.2,4){$\bf M $}\put(2.8, - .4){$\bf 5 $}\put(1, - .4){$\bf 5 $}\put(3.4, 2){$\bf 11 $}\put(2.3,0){\line(0,1){.3}}\put(2.3,.3){\line( - 1,0){.3}}\end{picture}

\underline{\blue{\sf In\: \triangle MPL \ \& \ \triangle MPN :- }}

\qquad \bullet LP = PN = 5 \:\:(given) \\\\\qquad \bullet MP = MP \:\:(Common) \\\\\qquad \bullet \angle MPN = \angle MPL = 90^{\circ} \:\: (given)

Hence by SAS congruence condition ,

\orange{\bf \triangle MPL \cong \triangle MPN }

Hence by cpct ( Corresponding parts of congruent triangles ) we can say that , LM = NM = 11 units .

<h3><u>Hence </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>LM</u><u> </u><u>is</u><u> </u><u>1</u><u>1</u><u> </u><u>units</u><u> </u><u>.</u></h3>
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