A car can travel 25 miles per gallon. Approximately how many gallons of fuel will the car need to travel 30 km?[1 mile = 1.6 km]

0.05
0.52
0.75
0.83

Answers

Answer 1
Answer:

Answer:

0.75 gallons of fuel will the car need to travel 30 km.

Step-by-step explanation:

As given

A car can travel 25 miles per gallon.

As given

1 mile = 1.6 km

Now convert 25 miles into km .

25 miles = 25 × 1.6 km

               = 40 km

i.e A car can travel 40 km per gallon.

40 km = 1 gallons.

1\ km = (1)/(40)\ gallons

1\ km = 0.025\ gallons

Now find out gallons of fuel will the car need to travel 30 km.

Than

gallons of fuel will the car need to travel 30 km = 30 × 0.025

                                                                                = 0.75 gallons

Therefore  0.75 gallons of fuel will the car need to travel 30 km.



Answer 2
Answer: 0.75 is the approximate answer

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Evaluate 5 - t/3 when t =12

Answers

Answer:

1

Step-by-step explanation:

Plug in t = 12 in the expression

5-(t)/(3)=5-(12)/(3)    

         = 5 - 4/1

          = 5 - 4

           = 1

Answer:

\Huge \boxed{1}

Step-by-step explanation:

\displaystyle 5-(t)/(3)

The value of t in the expression is 12.

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Find b, given that a = 18.2, B = 62°, and C = 48°. Round answers to the nearest whole number. Do not use a decimal point or extra spaces in the answer or it will be marked incorrect.

Answers

Answer:

17

Step-by-step explanation:

We have been given that in triangle ABC, measure of angle B is 62 degrees and measure of angle C is 48 degrees. The length of side opposite to angle a is 18.2. We are asked to find length of side b.

We will use law of sines to solve for side b.

\frac{a}{\text{sin}(A)}=\frac{b}{\text{sin}(B)}=\frac{c}{\text{sin}(C)}

m\angle A+m\angle B+m\angle C=180^(\circ)\n\nm\angle A+62^(\circ)+48^(\circ)=180^(\circ)

m\angle A+110^(\circ)=180^(\circ)

m\angle A+110^(\circ)-110^(\circ)=180^(\circ)-110^(\circ)

m\angle A=70^(\circ)

Upon substituting our given values, we will get:

\frac{18.2}{\text{sin}(70^(\circ))}=\frac{b}{\text{sin}(62^(\circ))}

\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))=\frac{b}{\text{sin}(62^(\circ))}*\text{sin}(62^(\circ))

\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))=b

b=\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))

b=(18.2)/(0.939692620786)*0.882947592859

b=19.1551999*0.882947592859

b=16.9130376

b\approx 17

Therefore, the length of side b is 17 units.

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Answers

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

2 x -3 = -6

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