1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bonufazy [111]
3 years ago
7

Just the answer Please hurry

Mathematics
2 answers:
Evgesh-ka [11]3 years ago
7 0

Step-by-step explanation:

476=14

578=X

578*14/476

8092/476=17

So the answer is 17 gallons

kaheart [24]3 years ago
5 0

Answer:

We have been given that a car can travel 476 miles on 14 gallons of gas and we are asked to write an equation relating the distance d to the number of gallons g.

We can set an equation as distance traveled is directly proportional to number of gallons used in travelling that distance.

Now we will find out K(number of miles traveled in one gallon of gas) by substituting our given values in this equation.

Therefore, we can see that car travels 34 miles in one gallon of gas.

Now we can use this information to find out how many gallons of gas does this car need to travel 578 miles.

Let us divide number of miles to be traveled by number of miles traveled in one gallon of gas to find out number of gallons of gas needed.

Therefore, car need 17 gallons of gas to travel 578 miles.

Step-by-step explanation:

You might be interested in
Solve this problem using 3.14 for pi. The diameter of this Frisbee is 12 inches. What’s the circumference of this Frisbee?
Ludmilka [50]
Circumference= diameter x pi
12x3.14 = 37.68
8 0
3 years ago
What is 0.080 as a fraction
nordsb [41]
That is 8/100 or 80/1000
3 0
3 years ago
Read 2 more answers
What is the missing numbers<br><br> 9 tenths + 7 hundredths<br><br> 6 tenths + _ hundredths = 0.66
dolphi86 [110]
The first one is 0.97 the second one is 6
7 0
3 years ago
Read 2 more answers
HJ
Olin [163]

Answer:

19 degrees

Step-by-step explanation:

From the question given

The interior angles are x+12 and x - 3

Exterior angle is <IJK = 5x-6

Using the rule that states that the sum of interior angle of a triangle is equal to the exterior

<JHI + <HIJ = <IJK

x+12 + x-3 = 5x - 6

2x+9 = 5x -6

2x - 5x = -6-9

-3x = -15

x = -15/-3

x = 5

Get <IJK

Recall that <IJK = 5x - 6

<IJK = 5(5) - 6

<IJK = 25-6

<IJK = 19 degrees

Hence the measure of <IJK is 19 degrees

8 0
3 years ago
Please help me find the total area of the composite figure below (geometry)
Akimi4 [234]

Answer:

lw + \frac{1}{2} × π × (\frac{l}{2} )^{2} ⇒ Answer D is correct

Step-by-step explanation:

First, let us find the area of the semi-circle.

Area = \frac{1}{2} × π × r²

<u>Given that,</u>

diameter of the semi-circle is ⇒ <em>l</em>

∴ radius ⇒ <em>l / 2</em>

<u>Let us find it now.</u>

Area = \frac{1}{2} × π × r²

Area =  \frac{1}{2} × π × (\frac{l}{2} )^{2}

<u>                                                     </u>

Secondly, let us find the area of the rectangle.

Area = length × width

<u>Given that,</u>

length ⇒ <em>l</em>

width ⇒ w

<u>Let us find it now.</u>

Area = length × width

Area = l ×w

Area = lw

<u>                                                      </u>

And now let us <u>find the total area.</u>

Total area =  Area of the rectangle + Area of the semi - circle

Tota area = lw + \frac{1}{2} × π × (\frac{l}{2} )^{2}

8 0
2 years ago
Other questions:
  • Are triangles ABC and EBC congruent?
    6·2 answers
  • I need help with number 22
    15·1 answer
  • Please help me with my algebra work.
    8·1 answer
  • I need help on the bottom problem
    8·2 answers
  • What is the the highest common factor of 25 and 45?
    14·1 answer
  • Please help
    13·1 answer
  • The endpoints of a line segments are at the coordinates (-6,3,4) and (4,-1,2) what is the midpoint of the segment
    15·1 answer
  • If 3v/7 = 6then v = ? <br> I’m not sure
    5·2 answers
  • Question 17 of 25
    7·1 answer
  • Solve for x, will give brainliest !!
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!