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3 years ago

10

With a full 12-gallon tank of fuel, Dillon’s car can travel about 420 miles. At that rate, about how far can Dillon’s car travel

on 5 gallons of fuel?

1 answer:

Xelga [282]3 years ago

3 0

175 miles~~~~~~~~~~~~

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Can anyone help me ?

taurus [48]

Answer:

_

66,6% = ⅔

_

33,3% = ⅓

85% = 17⁄20

80% = ⅘

75% = ¾

90% = 9⁄10

20% = ⅕

10% = ⅒

Step-by-step explanation:

To convert from a fraction to a percentage, divide both the denominator and numerator to get a decimal point, then move the decimal point twice to the right, then attach the percentage symbol:

⅒ = 0,1 = 10%

⤻⤻

⅕ = 0,2 = 20%

⤻⤻

9⁄10 = 0,9 = 90%

⤻⤻

¾ = 0,75 = 75%

⤻⤻

⅘ = 0,8 = 80%

⤻⤻

17⁄20 = 0,85 = 85%

⤻⤻

_ _

⅓ = 0,3 = 33,3%

⤻⤻

_ _

⅔ = 0,6 = 66,6

All those bars above those digits are what are known as bar notation, indicating that digits repeat.

I am joyous to assist you anytime.

6 0

3 years ago

12,13,14 help please

PSYCHO15rus [73]

12. 15 (3x5x1)
13. 8 (4x2x1)
14. 18 (6x3x1)

Hope this helps! :D

5 0

3 years ago

Read 2 more answers

Please awnser 1,2,3,and 4

Kryger [21]

1.B
2.e
3.8+10(x+3)+x
8+10x+30+x
11x=-30-8
x=38+11x

5 0

3 years ago

Find the HCF of 27, 63, 54 using prime factorization method.

Step2247 [10]

Answer:

27 = 3 , 3, 3

63= 3 , 3, 7

54= 3 , 3, 3, 2

HCF OF 27, 63 AND 54 = 3 X 3 = 9

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7 0

3 years ago

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago